Forrest Math148notes

8/21/2019 Forrest Math148notes

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Math 148: Calculus II

Instructor: Brian E. Forrest

December 30, 2009


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Chapter 1

Integration

This chapter is devoted to basic integral calculus; we explore the definitions,properties, and techniques of integration. We will prove the fundamental theo-rem of calculus —one of the most important theorems in analysis.

1.0 Areas Under a Curve

Problem. How do we find areas under the graph of a function f (x)?

Example 1.1. Determine the area bounded by the graph of f (x) = x2, thelines x = 0, x = 1, and y = 0.

1. Let A be the area of the region R.

2. We can approximate the area by a rectangle with height equal to f (1)(i.e., f (1) = 12 = 1) and the base over the interval [0, 1]. The area of thisrectangle is given by E 0 = height × base = 1 × 1 = 1. Note that A ≤ E 0.

3. Next we split the interval [0, 1] into two equal parts at x = 12 . In this case,R = R1 + R2 and we can estimate the area of R by estimating the areasof R1 and R2. Let A1 and A2 denote the area of R1 and R2, respectively.

Repeating these steps, we estimate A1 and A2 just as before—with rectanglesof heights, respectively, f ( 12) =

14 , f (1) = 1, and of bases, respectively,

12−0 = 12 ,

1 − 12 = 12 . Hence,

A1 ≈ f 121

2 − 0 = 1

221

2

= 18

and

A2 ≈ f (1)

1 − 12

= 12

1

2

=

1

2,

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whence A = A1 + A2 ≈ 18 + 12 = 58 .In general, we can divide the interval [0, 1] into n identical parts. Let Ai

denote the area of Ri; then

R = R1 + R2 + · · · + Rn,A = A1 + A2 + · · · + An.

We see that Ai ≈ f ( in)( in − i−1n ) = ( in)2 · 1n . Hence

A = A1 + · · · + An =n

i=1

Ai

≈n

i=1

( i

n)2 · 1

n

= 1n3

ni=1

i2

= 1

n3

n(n + 1)(2n + 1)

6

= 1 · (1 + 1n)(2 + 1n)

6 .

Note that limn→∞n

i=1( in)2( 1n) = limn→∞

1·(1+ 1n )(2+ 1n )6

= 13 .

Question: Is 13 the area of the region R?

We want A ≥ 13 , since this will show that A = 13 .

Strategy: Instead of taking the right-hand sum, we take the left-hand sum; weare now underestimating.

If we again divide [0, 1] into n equal parts to generate subregions R1, . . . Rn,we can once more estimate the area Ai of Ri by approximating Ri with rectanglewhose base is on [ i−1n ,

in ] and whose height equals f (

i−1n ). In this case, we have

Ai ≈ f

i − 1n

· 1

n =

i − 1

n

2· 1

n = (i − 1)2

1

n3

,

A ≥n

i=1

(i−1)2· 1n3

= 1

n3

n−1i=1

i2 = 1

n3

(n − 1)(n)(2n − 1)

6

=

(1 − 1n)(1)(2 − 1n )6

,

and hence

A ≥ limn→∞

(1 − 1n)(1)(2 − 1n )6

= 13

.

Conclusion: A = 13 . With this section as our motivation, we can now definewhat an integral is; as a result, we will be able to define formally the conceptof area .

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1.1 Integrals and Riemann Sums

Definition 1.1.1. Let [a, b] be a finite, closed interval. A partition P of [a, b]is a finite subset of [a, b] of the form

P = {xi : a = x0 < x1 < · · · < xn = b}.

Definition 1.1.2. For a partition P of [a, b] having n elements,1. define ∆xi := xi − xi−1, and2. define the norm of P , denoted as P, by

P := max{∆xi : i = 1, 2, . . . , n}.

Example 1.2. For any n ∈ N we can construct the n-regular partition P n of [a, b] by subdividing [a, b] into n identical parts with the length of each subin-terval being b−an . I.e., ∆xi =

b−an for each i, and P = b−an . Note that

x1 = a + b − a

n ,

x2 = a + 2 · b − an

,

...

xi = a + i · b − an

,

.

..xn = a + n · b − a

n = b.

Assume that f (x) is bounded on [a, b]. Let P be a partition on [a, b]. LetM i = sup{f (x) : x ∈ [xi−1, xi]},

and letmi = inf {f (x) : x ∈ [xi−1, xi]}.

Then mi ≤ f (x) ≤ M i for all x ∈ [xi−1, xi].

Definition 1.1.3. Given f (x), bounded on [a, b], and a partition P of [a, b], wedefine the upper Riemann sum of f (x) with respect to P by

U(f, P ) := Uba(f, P ) =n

i=1

M i · ∆xi,

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and the lower Riemann sum of f (x) with respect to P by

L(f, P ) := Lba(f, P ) =n

i=1

mi · ∆xi,

where M i and mi are defined as above. Finally, for each i, choose ci ∈ [xi−1, xi];a Riemann sum for f (x) on [a, b] with respect to P is defined by

Sba(f, P ) =n

i=1

f (ci)∆xi.

The bounds of summation a and b from definition 1.1.3 are usually omitted.

Remark 1.1.4. Since mi ≤

f (ci)

≤M i and ∆xi > 0, we can conclude that

Lba(f, P ) ≤ Sba(f, P ) ≤ Uba(f, P ).

Definition 1.1.5. Given a partition P , we say that a partition Q is a refinement of P if and only if P ⊆ Q. We also say that Q refines P or P is refined by Q.

Theorem 1.1.6. Let f (x) be bounded on [a, b]. Let P , Q be partitions on [a, b]with Q a refinement of P . Then

L(f,

P )≤

L(f,

Q)≤

U(f,

Q)≤

U(f,

P ).

Proof. Assume that Q = P ∪ {y0} for some y0 ∈ [a, b] \ P . Hence Q ={xi, y0 : a = x0 < x1 < · · · < xi−1 < y0 < xi < · · · < xn = b} andP = {xi : a = x0 < x1 < · · · < xn = b}. Let

M j = sup{f (x) : x ∈ [xj−1, xj ]},mi = inf {f (x) : x ∈ [xj−1, xj]}.

Also let

M i,1 = sup{f (x) : x ∈ [xi−1, y0]},mi,1 = inf {f (x) : x ∈ [xi−1, y0]},M i,2 = sup

{f (x) : x

∈[y0, xi]

},

mi,2 = inf {f (x) : x ∈ [y0, xi]}.

Note that since f ([xi−1, xi]) ⊇ f ([xi−1, y0]) and f ([xi−1, xi]) ⊇ f ([y0, xi]), we

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have that M i,1 ≤ M i and M i,2 ≤ M i. Similarly, mi,1 ≥ mi and mi,2 ≥ mi. Now

U(f, P ) = nj=1

M j · ∆xj

=nj=1j=i

+M j · ∆xj + M i · ∆xi

=nj=1j=i

+M j · ∆xj + M i(y0 − xi−1) + M i(xi − y0)

≥nj=1j=i

+M j · ∆xj + M i,1(y0 − xi−1) + M i,2(xi − y0)

= U(f, Q).Similarly,

L(f, P ) =n

j=1

mj · ∆xj

=nj=1j=i

+mj · ∆xj + mi · ∆xi

=nj=1j=i

+mj · ∆xj + mi(y0 − xi−1) + mi(xi − y0)

≤nj=1j=i

+mj · ∆xj + mi,1(y0 − xi−1) + mi,2(xi − y0)

= L(f, Q).

Due to remark 1.1.4, we conclude that

L(f, P ) ≤ L(f, Q) ≤ U(f, Q) ≤ U(f, P ).

We can use induction on the number of points in Q\P to establish the theorem.

Corollary 1.1.7. If P

and Q

are partitions of f (x) on [a, b], then L(f,P

)≤U(f, Q).

Proof. Let T = P ∪ Q. Then T refines both P and Q. Hence L(f, P ) ≤L(f, T ) ≤ U(f, T ) ≤ U(f, Q), as desired.

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Definition 1.1.8. Let f (x) be bounded on [a, b]. The upper Riemann integral for f (x) over [a, b] is b

a

f (x) dx = inf {U(f, P ) : P a partition of [a, b]},

and the lower Riemann integral for f (x) over [a, b] is ba

f (x) dx = sup{L(f, P ) : P a partition of [a, b]},

Observation. ba

f (x) dx ≥ ba

f (x) dx.

1.2 Integrable Functions

Definition 1.2.1. We say that f (x) is Riemann integrable on [a, b] if ba

f (x) dx =

ba

f (x) dx,

in which case we denote this common value by ba

f (x) dx.

In definitions 1.1.8 and 1.2.1, we encountered new notations: the is theintegral sign , a and b are endpoints of integration , f (x) is the integrand , and dxrefers to the variable of integration (also called the dummy variable )—in thiscase it is x.

Example 1.3. Let

f (x) =

1 if x ∈ [0, 1] ∩Q,−1 if x ∈ [0, 1] \ Q.

If P = {xi : 0 = x0 < x1 < · · · < xn = 1}, then letM i = sup{f (x) : x ∈ [xi−1, xi]},mi = inf {f (x) : x ∈ [xi−1, xi]}.

We have that

U(f, P ) =n

i=1

M i · ∆xi =n

i=1

∆xi = 1, and

L(f, P ) =n

i=1

mi · ∆xi =n

i=1

−∆xi = −1.

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Hence

10 f (x) dx = 1 = −1 =

1

0 f (x) dx,

and hence this function is not integrable on [0 , 1].

Observation. f (x) in the above example is discontinuous everywhere on [a, b].

Theorem 1.2.2. Let f (x) be bounded on [a, b]. Then f (x) is integrable on [a, b] if and only if for every > 0 there exists a partition P of [a, b] such that U(f, P ) − L(f, P ) < .Proof. Assume that f (x) is integrable on [a, b]. I.e.,

ba

f (x) dx =

ba

f (x) dx.

Let > 0. Since ba

f (x) dx = inf {U(f, P ) : P is a partition}, there exists apartition P 1 such that b

a

f (x) dx ≤ U(f, P 1) < ba

f (x) dx +

2.

Similarly, since

b

af (x) dx = sup{L(f, P ) : P is a partition}, there exists a par-

tition

P 2 such that b

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To prove the converse, assume that for each > 0 there exists a partition P of [a, b] such that

U(f, P ) − L(f, P ) < .However, by definition 1.1.8, we have that

L(f, P ) ≤ ba

f (x) dx ≤ ba

f (x) dx ≤ U(f, P ),

and hence

0 ≤ ba

f (x) dx − ba

f (x) dx < .

Since is arbitrary, we obtain

ba

f (x) dx = ba

f (x) dx;

therefore f (x) is integrable on [a, b], as required.

Example 1.4. Let f (x) = x2. Let P n be the n-regular partition of [0, 1]. Then

U10(f, P n) =n

i=1

f (xi)∆xi =n

i=1

i2

n2 · 1

n

and

L10(f,

P n) =

n

i=1 f (xi−1)∆xi =n

i=1(i − 1)2

n2

·

1

n

.

Then

U(f, P n) − L(f, P n) =

n2

n2 · 1

n

−

02

n2 · 1

n

=

1

n.

This shows that for all > 0, we can find a partition P such that U(f, P ) −L(f, P ) < (simply select a large enough n by the Archimedean Principle anduse the n-regular partition P n). Hence by theorem 1.2.2, f (x) is integrable on[0, 1]. Later, we will be able to show that

10

x2 dx = 13

.

Recall the following definition.

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Definition 1.2.3. We say that f (x) is uniformly continuous on an interval I if for every > 0, there exists a δ > 0 such that for all x, y

∈I ,

|x − y| < δ implies |f (x) − f (y)| < .Also recall the following theorem.

Theorem 1.2.4 [Sequential Characterization of Uniform Continu-ity]. A function f (x) is uniformly continuous on an interval I if and only if whenever {xn}, {yn} are sequences in I ,

limn→∞(xn − yn) = 0 implies limn→∞(f (xn) − f (yn)) = 0.

Example 1.5. Let I = R and f (x) = x2. Take

{xn

} = n + 1n, {yn} = {n}.Then limn→∞(xn − yn) = limn→∞ 1n = 0, but

limn→∞(f (xn) − f (yn)) = limn→∞

n +

1

n

2− n2

= limn→∞

2 +

1

n2

= 2 = 0.

Hence f (x) = x2 is not uniformly continuous on R.

One should already be familiar with the following theorem.

Theorem 1.2.5. If f (x) is continuous on [a, b], then f (x) is uniformly contin-uous on [a, b].

Proof. Assume that f (x) is continuous on [a, b] but not uniformly continuouson [a, b]. Then by theorem 1.2.4, there exists sequences {xn} and {yn} withlimn→∞(xn − yn) = 0 but limn→∞(f (xn) − f (yn)) = 0. (Note that this limitmay not even exist.) By choosing a subsequence if necessary, we can assumewithout loss of generality that there exists an 0 > 0 such that

(1.1) |(f (xn) − f (yn)) − 0| = |f (xn) − f (yn)| ≥ 0for all n ∈ N.

Since {xn} ⊂ [a, b], by the Bolzano-Weierstrass Theorem {xn} has a subse-quence

{xnk}

with limk→∞

xnk

= x0 ∈

[a, b]. Note that limk→∞

(xnk −

ynk

) = 0,hence limk→∞ ynk = x0 also. By the sequential characterization of continuity,we have

limk→∞

f (xnk) = f (x0),

limk→∞

f (ynk) = f (x0).

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Therefore, we can select a K ∈ N with

|f (xnk) − f (x0)| < 0

2 and

|f (ynk) − f (x0)| < 0

2

for all k ≥ K . Hence we have, for all k ≥ K ,0 ≤ |f (xnk) − f (ynk)| ≤ |f (xnk) − f (x0)| + |f (ynk) − f (x0)|

< 0

2 +

0

2= 0,

directly contradicting equation (1.1). Hence f (x) is uniformly continuous on[a, b].

Theorem 1.2.6 [Integrability Theorem for Continuous Functions].If f (x) is continuous on [a, b], then f (x) is integrable on [a, b].

Proof. Let > 0. Since f (x) is continuous on [a, b], f (x) is also uniformlycontinous on [a, b] by theorem 1.2.5 above. We can find a δ > 0 such thatif |x − y| < δ , then |f (x) − f (y)| < b−a . Let P be a partition of [a, b] withP = max{∆xi} < δ . For example, we can choose the n-regular partitionof [a, b] with n large enough (chosen by the Archimedean Principle) so thatb−an = P n < δ . After we have chosen P , let

M i = sup{f (x) : x ∈ [xi−1, xi]},mi = inf {f (x) : x ∈ [xi−1, xi]}.

By the extreme value theorem, there exists ci, di ∈ [xi−1, xi] such that f (ci) =mi and f (di) = M i; note that M i − mi = |M i − mi| = |f (di) − f (ci)|. Since|ci − di|


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Remark 1.2.7. Assume that f (x) is continuous on [a, b]. For each n, let P n bethe n-regular partition. Then if S(f,

P n) is any Riemann sum associated with

P n, we have ba

f (x) dx = limn→∞S(f, P n).

Proof. Given > 0, we can find an N large enough so that if n ≥ N , thenb−an < δ as defined in theorem 1.2.6 above. Hence

U(f, P n) − L(f, P n) < ,via a similar argument made in theorem 1.2.6. But U(f, P n) ≥ S(f, P n) ≥L(f, P n) and U(f, P n) ≥

ba

f (x) dx ≥ L(f, P n). This implies that

b

a

f (x) dx − S(f, P n) < and so limn→∞ S(f, P n) =

ba

f (x) dx, as remarked.

Note that the above remark shows that if f (x) is continuous, we have ba

f (x) dx = limn→∞

ni=1

f

a + i · b − a

n

b − a

n

.

Definition 1.2.8. Given a partition P = {xi : a = x0 < x1 0, there

exists δ > 0 such that if P is a partition of [a, b] with P < δ , then S(f, P ) − ba

f (x) dx

< for any Riemann sum S(f, P ).

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Proof. Since f (x) is integrable, we can find a partition P 1 of [a, b] with U(f, P 1)−L(f,

P 1) <

2 . In particular, for any Riemann sum S(f,

P 1), we have U(f,

P 1)

≥S(f, P 1) ≥ L(f, P 1) and U(f, P 1) ≥ ba f (x) dx ≥ L(f, P 1). Suppose P 1 ={xi : a = x0 < x1 < · · · < xn = b}. Then letM = sup{f (x) : x ∈ [a, b]},m = inf {f (x) : x ∈ [a, b]}.

If M −m = 0, then f (x) is constant on [a, b], so say f (x) = c for all x ∈ [a, b].In this case U(f, P ) = c(b − a) = L(f, P ) for any partition P . Hence any δ > 0would satisfy the theorem and we are done. So assume M > m.

Let δ < 2n(M −m) . Let P = {yi : a = y0 < y1 < · · · < yj < · · · < yk = b} beany partition of [a, b] with P < δ . Let

T = { j : j ∈ {1, 2, . . . , k}, and [yj−1, yj ] ⊆ [xi−1, xi] for some i}.Also, let

M j = sup{f (x) : x ∈ [yj−1, yj ]},mj = inf {f (x) : x ∈ [yj−1, yj]}.

Then we have that

U(f, P ) − L(f, P ) =k

j=1

M j · ∆yj −k

j=1

mj · ∆yj

=k

j=1

(M j − mj)∆yj

= j∈T(M

j − mj)∆yj + j∈{1,2,...,k}\T(M

j − mj)∆yj .

Note that j∈T

(M j − mj)∆yj ≤ U(f, P 1 ∪ P ) − L(f, P 1 ∪ P )

≤ U(f, P 1) − L(f, P 1)<

2.

Observe that {1, 2, . . . , k} \ T has at most n elements (since for each j ∈ T, wehave a unique i such that yj−1 < xi−1 < yj < xi; conversely for each such ithere exists a unique j ∈ T—but there are only n points in P 1). Hence for each

j ∈ {0, 1, . . . , k} \ T, we have(M j − mj)∆yj ≤ (M − m) ·P

< (M − m) · 2n(M − m)

=

2n.

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This shows that

j∈{1,2,...,k}\T

(M j − mj)∆yj < j∈{1,2,...,k}\T

2n

≤ 2n

· n

=

2.

Hence, if P < δ , U(f, P ) −L(f, P ) < 2 + 2 = ; therefore, if S(f, P ) is anyRiemann sum, then S(f, P ) −

ba

f (x) dx

< ,as required.

This thoerem is a generalization of remark 1.2.7. In particular, this theoremprovides an easy, alternate proof of remark 1.2.7. Here we state the remarkagain as a corollary.

Corollary 1.2.10. If f (x) is integrable on [a, b], then ba

f (x) dx = limn→∞S(

f, P n) = limn→∞

ni=1

f (ci) · b − an

,

where P n is the n-regular partition of [a, b] and ci ∈ [xi−1, xi]. ( S(f, P n) is any Riemann sum.)

Proof. Let > 0. By the above theorem, we can find δ > 0 so that if P n =1n < δ , S(f, P n) −

ba

f (x) dx

< for any Riemann sum S(f, P n). The result follows by choosing an N (using theArchimedean Principle) so that 1n ≤ 1N < δ for all n ≥ N .

In general, we write ba

f (x) dx = limP→0

S(f, P ).

Theorem 1.2.11. Let f (x) be monotonic on [a, b]. Then f (x) is integrable on [a, b].

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Proof. Let P n be the n-regular partition. Assume, without loss of generality,that f (x) is nondecreasing on [a, b]. Then

U(f, P n) = SR(f, P n) =n

i=1

f (xi) · b − an

, and

L(f, P n) = SL(f, P n) =n

i=1

f (xi−1) · b − an

.

So

U(f, P n) − L(f, P n) =n

i=1

f (xi) · b − an

−n−1i=0

f (xi) · b − an

= b − a

n

[(f (x1) +

· · ·+ f (xn))

−(f (x0) + · · · + f (xn−1))]=

b − an

(f (xn) − f (x0))

= b − a

n (f (b) − f (a)).

Since limn→∞ b−an (f (b) − f (a)) = 0, f (x) is integrable on [a, b] by theorem1.2.2.

Question: Assume that f (x) is continuous on [a, b] except at c. (Note thatf (x) is bounded on [a, b].) Is f (x) still integrable on [a, b]?

Theorem 1.2.12. Assume that f (x) is continuous on [a, b] except at c ∈ [a, b].Then f (x) is integrable on [a, b].

Proof. Let P = {xi : a = x0 < x1 0. Let

M = sup{f (x) : x ∈ [a, b]},m = inf {f (x) : x ∈ [a, b]},

M i = sup{f (x) : x ∈ [xi−1, xi]},mi = inf {f (x) : x ∈ [xi−1, xi]}.

We can choose P so that ∆xi0 < 3(M −m) . Then (M i0 − mi0)∆xi0 ≤(M

−m)∆xi0 < (M

−m) 3(M

−m) =

3 . Since f (x) is uniformly continuous

on [a, xi0−1], by refining as necessary, we can assume that if 0 ≤ i ≤ i0, thenM i − mi < 3(xi0−a) . We now have

i0−1i=1

(M i − mi)∆xi <i0−1i=1

3(xi0−1 − a)∆xi =

3

i0−1i=1

∆xixi0−1 − a

=

3.

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A similar argument shows, by refining if necessary, that

ni=i0+1

(M i − mi)∆xi < 3

.

Then

U(f, P ) − L(f, P ) =n

i=1

(M i − mi)∆xi

=

i0−1i=1

(M i − mi)∆xi + (M i0 − mi0)∆xi0

+

n

i=i0+1(M i − mi)∆xi

<

3 +

3 +

3= .

This shows that f (x) is integrable on [a, b] by theorem 1.2.2.

Observation. Observe the following facts:

1. If f (x) is bounded on [a, b] and continuous except at possibly at finitelymany points, then f (x) is integrable on [a, b].

2. If f (x) is integrable on [a, b] and g(x) = f (x) except at finitely manypoints, then g (x) is integrable and b

a

f (x) dx =

ba

g(x) dx.

In fact, much more is true. Let (a, b) be an open interval. Define the length of (a, b) to be l(a, b) = b −a. We say that a subset A ⊂ R has Lebesgue measure zero if for each > 0, there exists a sequence {(an, bn)} of open intervals suchthat

A ⊆∞i=1

(ai, bi) and

∞i=1

l(ai, bi) < .

Diversion. Suppose A ⊂ R is countable. Then A has Lebesgue measure zero.Proof. Assume that A = {x1, x2, . . . , xi, . . .} is countable. Let > 0. Let

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(an, bn) = (x − 2n+1 , x + 2n+1 ). Clearly, A ⊂∞i=1(an, bn). Also,

∞i=1

l(an, bn) =∞i=1

l(x − 2n+1

, x +

2n+1)

=∞i=1

x +

2n+1 − x +

2n+1

=

∞i=1

2n

= ,

implying that A has Lebesgue measure zero.

Diversion. Let f (x) be a function on [a, b]. If the set of points of discontinuityof f (x) on [a, b] has Lebesgue measure zero, then f (x) is integrable on [a, b].

1.3 Properties of Integrals

Theorem 1.3.1. Assume that f (x) is integrable on [a, b], then

1. (cf )(x) = c · f (x) is integrable on [a, b] and ba (cf )(x) dx = c ba f (x) dx for all c ∈ R.

2. If g(x) is also integrable on [a, b], then (f +g)(x) = f (x)+g(x) is integrable

on [a, b] and ba (f + g)(x) dx = ba f (x) dx + ba g (x) dx.Proof. To prove (1), we consider three cases: c = 0, c > 0, and c 0. Since f (x) is integrable, we can choose(by theorem 1.2.2) a partition P such that Uba(f, P ) − Lba(f, P ) < |c| . It is clearthat

U(cf, P ) = c · U(f, P ), andL(cf,

P ) = c

·L(f,

P ),

if c > 0 and

U(cf, P ) = c · L(f, P ), andL(cf, P ) = c · U(f, P ),

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if c


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Theorem 1.3.4. Suppose f (x) is integrable on [a, b]. Then

ba f (x) dx ≤ ba |f (x)| dx.Proof. ¿From the observation above, we have that b

a

f (x) dx =

ba

[f +(x) − f −(x)] dx

=

ba

f +(x) dx − ba

f −(x) dx (by theorem 1.3.1),

and that

b

a

|f (x)| dx = b

a

[f +(x) + f −(x)] dx

=

ba

f +(x) dx +

ba

f −(x) dx (by theorem 1.3.1).

Since f +(x) ≥ 0 and f −(x) ≥ 0 on [a, b], ba

f +(x) dx ≥ 0 and ba

f −(x) dx ≥ 0(all Riemann sums Sba(f +, P ) and Sba(f −, P ) are greater than or equal to 0 forall partitions P of [a, b]). Hence we can now show that

ba

f (x) dx

= ba

f +(x) dx − ba

f −(x) dx

≤

ba

f +(x) dx

+

ba

f −(x) dx

= ba

f +(x) dx + ba

f −(x) dx

=

ba

|f (x)| dx,

as required.

Definition 1.3.5 [Geometric Interpretation of the Definite Inte-gral]. If f (x) ≥ 0 on [a, b] and integrable on [a, b], then the area under the curve bounded by y = f (x), y = 0, x = a and x = b is

b

a

f (x) dx.

Now suppose f (x) is no longer restricted to be nonnegative on [a, b]. Then thearea above the x-axis bounded by y = f (x), y = 0, x = a and x = b is b

a

f +(x) dx.

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Similarly, the area below the x-axis bounded by y = f (x), y = 0, x = a and x = b is b

a

f −(x) dx.

¿From these, we can finally define the area bounded by y = f (x), y = 0, x = aand x = b, A, as the sum of the area above the x-axis and the area below thex-axis bounded by the same curves; i.e.,

A :=

ba

f +(x) dx +

ba

f −(x) dx = ba

|f (x)| dx.

Theorem 1.3.6. Let f (x) be bounded on [a, b] and let a < c < b. Then f (x) is integrable on [a, b] if and only if f (x) is integrable on [a, c] and on [c, b]. In this case, we have b

a

f (x) dx = ca

f (x) dx + bc

f (x) dx.

Proof. Assume that f (x) is integrable on [a, b]. Let > 0. Since f (x) is inte-grable, we can find a partition P with

P = {xi : a = x0 < x1 < · · · < xn = b}for which

U(f, P ) − L(f, P ) < .By refining if necessary, we can assume that c = xk for some 0 < k < n, sosuppose that

P =

{xi : a = x0 < x1 <

· · ·< xk = c <

· · ·< xn = b

}.

Now let P 1 = {xi : a = x0 < x1


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This shows that f (x) is integrable on [a, c] and [c, b] respectively, by theorem1.2.2.

Now assume the converse: f (x) is integrable on [a, c] and on [a, b]. Let > 0.Let P 1 = {xi : a = x0 < x1


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Remark 1.3.8. All of our properties of integrals are still valid with this ex-panded definition.

For example, we can show that the following theorem is true.

Theorem 1.3.9. Let f (x) be bounded and integrable on an interval I containing a and b, a = b. Suppose that m ≤ f (x) ≤ M for all x ∈ I . Then

m ≤ 1b − a

ba

f (x) dx ≤ M.

Proof. We know that if a < b, then the theorem holds. Indeed, if P = {xi : a =x0 < x1 < . . . < xn = b} is any partition of [a, b] and Lba(f, P ) is any lowerRiemann sum, then we have

m(b − a) = m ni=1

∆xi

=n

i=1

m · ∆xi

≤n

i=1

mi · ∆xi

= Lba(f, P )

≤ ba

f (x) dx,

where mi := inf f ([xi−1, xi]).

Similarly, ba f (x) dx ≤ M (b−a). Dividing these two inequalities by the positive

quantity b − a yields the theorem.To prove the theorem for the case when a > b, we simply note that

1

b − a ba

f (x) dx = 1

a − b · − ba

f (x) dx = 1

a − b ab

f (x) dx

by definition 1.3.7. We now apply the first paragraph of the proof to thisequivalent quantity with the roles of a and b reversed and the theorem follows.

Definition 1.3.10. If f (x) is integrable on [a, b], then we define the average value or the mean value of f on [a, b] as the quantity

1

b − a ba

f (x) dx.

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We now give two intuitive justifications for the terminology average value indefinition 1.3.10.

1. Geometric justification. The “signed area” spanned by the curve of f over [a, b] indicates the “sum” of all values of f over [a, b]. Dividing thisby the length of the interval [a, b] indicates, on average, how much eachvalue of f (x) contributed to the area:

1

b − a ba

f (x) dx.

That is to say, the average value µ of f should accomplish the following:The area of the rectangle whose base is [a, b] and whose height is µ is equalto the area of the region bounded by f whose base is also [a, b]. One canview µ as the average height of f on [a, b].

2. Sampling. We can sample the values of f at different points and simplyaverage them (in the finite sense). Then we can let the number of samplepoints go to infinity (in a way that we end up sampling each part of the curve equally). To accomplish this, let P n be the n-regular partitionof [a, b]. The average of the values of the right-hand endpoints of thispartition is:

f (x1) + f (x2) + · · · + f (xn)n

= 1

b − a

f (x1)b − a

n + f (x2)

b − an

+ · · · + f (xn) b − an

=

1

b − an

i=1f (xi)∆xi

= 1

b − a SR(f, P n).

Letting n → ∞—i.e., letting the number of sample points go to infinityin a way that different parts of the curve get sampled equally—we obtain

limn→∞

1

b − a SR(f, P n) = 1

b − a ba

f (x) dx,

by theorem 1.2.10.

1.4 The Fundamental Theorem of CalculusIn this section we will prove an important theorem: the fundamental theoremof calculus. First we will prove something about integral functions—functionsof the form F (x) =

xa

f (t) dt.

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Theorem 1.4.1. Assume that f (t) is defined on an interval I containing a point a and for each x

∈I , the integral xa f (t) dt exists. Also assume that there exists an M ∈ R such that |f (t)| ≤ M , for all t ∈ I . Then the function

F (x) :=

xa

f (t) dt

satisfies |F (x) − F (y)| ≤ M |x − y|

for all x, y ∈ I .Proof. First note that

ψφ

f (t) dt

=

φψ

f (t) dt

.

for all φ, ψ ∈ I . With this observation, we have, with the help of theorems 1.3.4and 1.3.6, that

|F (x) − F (y)| = xa

f (t) dt − ya

f (t) dt

=

xa

f (t) dt +

ay

f (t) dt

=

xy

f (t) dt

≤

x

y

|f (t)| dt

≤ x

y

M dt

= M |x − y|.The second inequality follows from an exercise in Riemann sums.

Corollary 1.4.2. With notation the same as the above theorem, F (x) is uniformly continuous on I .

Proof. Let > 0. Put δ = M . Then if x, y ∈ I are such that |x − y| < δ , thenwe have

|F (x)

−F (y)

| ≤M

|x

−y

|< M δ = M

M = ,

implying the theorem.

Question: Is it possible to show that F (x) is also differentiable? If so, what isF (x)?

This major theorem answers the above question.

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Theorem 1.4.3 [First Fundamental Theorem of Calculus]. Assume that f (t) is defined on an interval I containing a. Also suppose that xa f (t) dtexists for all x ∈ I ; define

F (x) :=

xa

f (t) dt.

Let x0 be in the interior of I (i.e., there exists an > 0 such that (x0−, x0+) ⊂I ). If f (t) is continuous at x0, then F (x) is differentiable at x0 and

F (x0) = f (x0).

Proof. Consider

F (x0 + h)

−F (x0) =

x0+h

a

f (t) dt

− x0

a

f (t) dt

=

x0+hx0

f (t) dt.

Given > 0, since f (t) is continuous at x0, we can find a δ > 0 such that if |t − x0| < δ , then |f (t) − f (x0)| < . Hence for each t ∈ (x0− δ, x0+ δ ), we have

(1.2) f (x0) − < f (t) < f (x0) + .

Let h be such that 0 < |h| < δ . First note that,

F (x0 + h) − F (x0)h

= 1

h

x0+hx0

f (t) dt

and so by inequality (1.2) and theorem 1.3.9, we have that

(1.3) f (x0) − ≤ F (x0 + h) − F (x0)h

≤ f (x0) + .

For all h satisfying 0 < |h| < δ , we can combine inequality (1.3) with inequality(1.2) (with t = x0) to obtainF (x0 + h) − F (x0)h − f (x0)

< .This implies that

F (x0) = limh→0 F (x

0 + h) − F (x0)h = f (x0)

and the proof is complete.

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Example 1.6. Let R+ = {x ∈ R : x > 0} = (0, ∞). Let f : R+ → R be definedby f (t) = 1t . Also, let F : R

+

→R be defined by

F (x) =

x1

f (t) dt =

x1

1

t dt.

The first fundamental theorem of calculus shows that F (x) = 1x for all x > 0.We know that if H (x) = ln(x), then H (x) = 1x for all x > 0. Let G(x) =F (x) − H (x). Then

G(x) = F (x) − H (x) = 1x − 1

x = 0.

The mean value theorem shows that there exists a C ∈ R such that G(x) = C for all x > 0; however,

C = G(1) := F (1) − H (1) = 11

1t

dt − ln(1) = 0 − 0 = 0.

We conclude that

ln(x) =

x0

1

t dt

for all x ∈ R+.

Example 1.7. Evaluate ddx x0

et2

dt.

Note that et2

is continuous on R, and so the first fundamental theorem of calculus shows that

d

dx x

0 e

t2

dt = e

x2

for all x ∈ R.

Example 1.8. Evaluate F (x), where

F (x) :=

x20

sin(t3) dt.

Let

H (u) :=

u0

sin(t3) dt,

then F (x) = H (x2), whence by the chain rule, F (x) = H (x2) ddx

x2 = 2xH (x2).By the first fundamental theorem of calculus, H (u) = sin(u3), and henceH (x2) = sin(x6). We conclude that

F (x) = 2xH (x2) = 2x sin(x6).

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Example 1.9. Let F : R → R be defined by

F (x) = x2x

sin(t3) dt.

Find F (x).Note that

F (x) =

x2x

sin(t3) dt

=

x2π

sin(t3) dt +

πx

sin(t3) dt

= x2

π

sin(t3) dt − x

π

sin(t3) dt,

and so by the fundamental theorem of calculus and the previous example, wehave that

F (x) = 2x sin(x6) − sin(x3).

As seen from the previous examples, the first fundamental theorem of cal-culus easily lends itself to the following generalization.

Corollary 1.4.4. Assume that f (t), g(x), and h(x) is defined on an interval I . Also suppose that g(x) and h(x) are both differentiable on the interior of I ,

and that h(x)g(x) f (t) dt exists for all x ∈ I . Define F : I → R by F (x) :=

h(x)g(x)

f (t) dt.

Let x0 be in the interior of I (i.e., there exists an > 0 such that (x0−, x0+) ⊂I ). If f (t) is continuous at x0, then F (x) is differentiable at x0 and

F (x0) = f (h(x0))h(x0) − f (g(x0))g(x0).

Proof. This proof is left as an exercise. (Hint: Mimic the procedure in theprevious examples.)

Recall the following definition.

Definition 1.4.5. Suppose that f (x) is defined on an open interval containingan interval I . We say that F (x) is an antiderivative of f (x) on I if F (x) = f (x)for all x ∈ I .

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Example 1.10. If f (x) = x, then F (x) = 12x2 is an antiderivative on of f (x)

on R.

Observation. Recall that if F (x), G(x) are continuous on an interval [a, b] andare antiderivatives of f (x) on (a, b), where a < b, then H (x) := F (x) − G(x) issuch that H (x) = F (x) − G(x) for all x ∈ (a, b). Since H (x) is continuous on[a, b], the mean value theorem shows that H (x) = C ∈ R is a constant for allx ∈ [a, b]; therefore, all antiderivatives of f (x) on (a, b) are of the form

F (x) + C.

Conversely, all such functions are antiderivatives of f (x) on (a, b). Hence if wefind one antiderivative of f (x) on (a, b), we have found all of them. In general,if F (x) is an antiderivative of f (x) on (a, b), we write

(1.4)

f (x) dx = F (x) + C.

The integral on the left side of (1.4) is called an indefinite integral andrepresents the family of antiderivatives of f (x) on (a, b). The f (x) is, as before,called the integrand and the dx refers to the variable of (indefinite) integration .The following theorem, the second fundamental theorem of calculus , relates theindefinite integral with the definite integral and explains their similar notation.

Theorem 1.4.6 [Second Fundamental Theorem of Calculus]. Assume that f (x) is defined on an interval [a, b], a < b. Assume that f (x) is continuous on [a, b] and differentiable on (a, b). If G(x) is continuous on [a, b] and is an antiderivative of f (x) on (a, b), then b

a

f (x) dx = G(b) − G(a).

Proof. First note that f (x) is integrable on [a, b] by theorem 1.2.6. So let F :[a, b] → R be defined by

F (x) =

xa

f (t) dt.

The first fundamental theorem of calculus shows that F (x) = f (x) on (a, b);i.e., F (x) is an antiderivative of f (x) on (a, b). In observation 5, we noted (by

the mean value theorem) that G(x) = F (x) + C on [a, b] for some C ∈ R. We

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now have

G(b) − G(a) = [F (b) + C ] − [F (a) + C ]= F (b) − F (a)

:=

ba

f (x) dx − aa

f (x) dx

=

ba

f (x) dx − 0

=

ba

f (x) dx,

as required.

Definition 1.4.7. Suppose f : S → R with S ⊆ R. If a, b ∈ S , then we definethe quantity f (x)|ba as follows:

f (x)|ba := f (b) − f (a).With this definition, the second fundamental theorem of calculus can be

rewritten as follows: ba

f (x) dx =

f (x) dx

ba

.

The second fundamental theorem allows us to compute exact numeric valuesfor certain definite integrals. Let us return to the first problem posed in thefirst section: What is the area bounded by the curve y = x2 and the lines y = 0,x = 0, x = 1?

Problem. Find 10

x2 dx.

Solution. Note that

x2 dx = 13x3 + C is an antiderivative. Let G(x) = 13x

3.By the second fundamental theorem of calculus, we have that 1

0

x2 dx = G(x)|10 := G(1) − G(0) = 1

3 − 0 = 1

3,

as expected.

Remark 1.4.8. Here is a list of common antiderivatives. The integrands aredefined on their natural domains and the antiderivatives are valid on thosedomains.

1.

c dx = cx + C ,

2.

xα dx = 1α+1xα+1 + C for all α = −1,

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3.

x−1 dx =

1x dx = ln(|x|) + C ,

4. sin(x) dx = − cos(x) + C ,5.

cos(x) dx = sin(x) + C ,

6.

ex dx = ex + C ,

7. 1√ 1−x2 dx = arcsin(x) + C ,

8. −1√ 1−x2 dx = arccos(x) + C ,

9.

sec2(x) dx = tan(x) + C ,

10.

tan(x) dx = ln(| sec(x)|) + C ,11. ln(x) dx = x(ln(x) − 1) + C .Proof. We will check the last two items and leave the rest as exercises.10. Note that

d

dx [ln(| sec(x)|) + C ] = 1| sec(x)| ·

d

dx| sec(x)|

+ 0

= 1

| sec(x)| ·

d

dx| sec(x)|

by the chain rule. For x ∈ −π2 + 2nπ, π2 + 2nπ (n any natural number),sec(x) > 0, and so

1

| sec(x)| · d

dx | sec(x)| = 1

sec(x) · d

dx sec(x)

= 1

sec(x) · sec(x) tan(x)

= tan(x),

as claimed. The case where x ∈ π2 + 2nπ, 3π2 + 2nπ is similar.11. Note that by the product rule,

d

dx [x ln(x) − x + c] = 1 · ln(x) + x · 1

x − 1 + 0 = ln(x) + 1 − 1 = ln(x)

for all x > 0, as required.

Note that indefinite integration satisfy the linearity property just as deriva-tives do. That is,

f (x) + g(x) dx =

f (x) dx +

g(x) dx and

c · f (x) dx =

c · f (x) dx, for all continuous f (x), g(x) and c ∈ R. This is due to the linearityof the derivative.

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1.5 Substitution and the Change of Variables

Theorem1.5.1 Substitution

Example 1.11. Evaluate

xex2

dx.

Observe that

xex2

dx = 12

(2x)ex2

dx = 12(ex2 + C 0) =

12e

x2 + C , where

C = 12C 0. To see this, note that the chain rule shows ddx12

ex2

+ C = 12 · 2xex2

=

xex2

.

Observation. Suppose that f (x) and g(x) are defined on an open interval I and differentiable on I . Recall that the chain rule states

d

dx f (g(x)) = f (g(x))g(x),

valid on I . Now we have that d

dx f (g(x)) dx =

f (g(x))g(x) dx

and hence

f (g(x)) + C =

f (g(x))g(x) dx.

Also note that f (u) du = f (u) + C.

Let u

= g

(x

); we get that f (g(x))g(x) dx = f (g(x)) + C = f (u) + C = f (u) + C =

f (u) du.

This suggests that if u = g(x), then “du = g(x) dx.”

In this section, we will attempt to find antiderivatives by this technique,known as (indefinite) integration by substitution . It is, in essence, the chain rulebackwards.

Sample Question 1. Evaluate

2xex2

dx.

Strategy: We want to put this integral into the form of f (g(x))g(x) dx.To do this, we try to match each factor of 2xex2 to a factor of f (g(x))g(x).The most promising match looks like f (u) = eu with u = g(x) := x2, whereg(x) = 2x. In other words, we try to find a quantity in the variable x—callit u—such that its differential · · · dx absorbs most of the difficult parts of theintegrand away into du; after this “absorption,” what remains of the integral

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should be expressible in terms of the quantity u only. With luck, whateverremains is easier to integrate.

In this question, if we choose our magic quantity to be u = x2, then noticethat du = 2x dx absorbs a large chunk of the integrand away into du, so thatwhat remains is ex

2

du. Now ex2

is just eu, and so the integral turns into eu du, a trivial integral. Usually, this magic quantity u can be found inside

the integrand itself; in this case, it is found in the exponent of e. Do not forget tore-express the result in terms of the original variable by substituting the magicquantity u back into the result.

Sample Solution 1. Let u = x2. We get that du = 2x dx. After substitutingu = x2 and du = 2x dx into

2xex

2

dx, we get that

2xex2 dx = eu du = eu + C.We can now substitute u = x2 back into our new antiderivative to get

2xex2

dx = eu + C = ex2

+ C.


tan(x) dx := sin(x)cos(x)

dx. Let u = cos(x), whence

du = − sin(x) dx, or du− sin(x) = dx. Now we have that

sin(x)cos(x) dx = sin(x)u · du− sin(x)= −

1

u du

= −(ln(|u|) + C 0)= − ln(|u|) + C (where C = −C 0)= ln

1

|u|

+ C

= ln

1cos(x)

+ C

= ln(| sec(x)|) + C.


xx+1 dx.

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Note that

xx + 1 dx =

x + 1x + 1 dx − 1x + 1 dx

=

1 dx −

1

x + 1 dx.

We know that

1 dx = x + C 0. Now let u = x + 1, whence du = dx and 1

x + 1 dx =

1

u du = ln(|u|) + C = ln(|x + 1|) + C 1.

We conclude that x

x + 1 dx = (x + C 0) − (ln(|x + 1|) + C 1) = x − ln(|x + 1|) + C,

where C = C 0 + C 1.

An alternate solution is to let u = x + 1 right away to obtain x

x + 1 dx =

u − 1

u du

=

1 − 1

u

du

=

1 du −

1

u du

= (u + C 0) − (ln(|u|) + C 1)= x − ln(|x + 1|) + 1 + C 0 − C 1= x − ln(|x + 1|) + C,

where C = 1 + C 0 − C 1.


sec(θ) dθ.We make the not-so-trivial observation that

sec(θ) dθ =

sec(θ)

sec(θ) + tan(θ)

sec(θ) + tan(θ)

dθ

=

sec2(θ) + sec(θ)tan(θ)

sec(θ) + tan(θ) dθ.(1.5)

We make the substitution u = sec(θ) + tan(θ), and so du = [sec(θ)tan(θ) +sec2(θ)] dθ. Now note that the entire numerator of 1.5 is absorbed into du andthe entire denominator becomes u, so that we obtain

sec(θ) dθ = sec2(θ) + sec(θ) tan(θ)

sec(θ) + tan(θ) dθ.

=

1

u du

= ln(|u|) + C = ln(| sec(θ) + tan(θ)|) + C.

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Example 1.15. Let f : [−1, 1] → R be defined by f (x) = √ 1 − x2. Evaluate f (x) dx =

1 − x2 dx,

valid on [−1, 1].The solution to this type of problem involves an inverse trigonometric sub-

stitution , which will be explored in more detail in a later section.Let θ = arcsin(x). Since x ∈ [−1, 1], θ ∈ −π2 , π2 and so cos(θ) ≥ 0 on

this interval. We have that x = sin(θ), and so dx = cos(θ) dθ. We now have

that√

1 − x2 =

1 − sin2(θ) =

cos2(θ) = | cos(θ)| = cos(θ). Putting thesetogether, we have

1 − x2 dx =

cos(θ) cos(θ) dθ

=

cos2(θ) dθ

=

cos(2θ) + 1

2

= 1

2

cos(2θ) dθ +

1

2

1 dθ

= 1

2 · sin(2θ)

2 +

1

2θ + C

= 1

4

sin(2θ) + θ

2

+ C

= 1

4 sin(2 arcsin(x)) +

arcsin(x)

2 + C.

Checking, we have

d

dx

1


arcsin(x)

2 + C

=

1

4 cos(2 arcsin(x)) · 2√

1 − x2 + 1

2√

1 − x2=

1

2√

1 − x2 (cos(2 arcsin(x)) + 1) dθ

= 1

2

√ 1 − x2 2cos

2(arcsin(x))

−1 + 1

= 1√

1 − x2 cos2(arcsin(x)).(1.6)

But observe that

cos2(arcsin(x)) + x2 = cos2(arcsin(x)) + sin2(arcsin(x)) = 1,

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and socos2(arcsin(x)) = 1

−x2.

Continuing from (1.6) above, we have that

d

dx

1


arcsin(x)

2 + C

=

1√ 1 − x2 · (1 − x

2)

=

1 − x2,and we are done.

1.5.2 Change of Variables Theorem

How does substitution work for definite integrals? What happens to the boundsof integration after we attempt to make a substitution?

Definition 1.5.1. Suppose g(x) is defined on an interval [a, b]. We say thatg(x) is continuously differentiable on [a, b] if g(x) is differentiable on (a, b), g (x)is continuous on (a, b), and the one-sided limits limx→a+ g(x) and limx→b− g(x)exist.

Theorem 1.5.2 [Change of Variables]. Assume that g(x) is defined and continuous on an interval [a, b], a < b. Suppose also that g(x) is continuously differentiable on [a, b]. Suppose also that f (x) is defined and continuous on g([a, b]). Then

b

a

f (g(x))g(x) dx = g(b)

g(a)

f (u) du,

where g(x) has been extended continuously to [a, b].

Proof. Note that g([a, b]) is a closed, bounded interval by the extreme valuetheorem. Also note that theorem 1.2.6 shows that f (u) is integrable on [a, b];let F : I → R be defined by

F (t) =

tg(a)

f (u) du.

By the first fundamental theorem of calculus, F (t) is differentiable on I ⊃g([a, b]) with F (t) = f (t) on I ⊃ g([a, b]). Let H : [a, b] → R be defined by

H (x) = F (g(x)) = g(x)

g(a)

f (u) du.

Since g(x) is differentiable on (a, b), the chain rule shows that for all x ∈ (a, b),we have that

H (x) = F (g(x))g(x)= f (g(x))g(x).

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That is, H (x) is an antiderivative of f (g(x))g(x) on (a, b). Extend g(x) con-tinuously to [a, b] so that f (g(x))g(x) is now continuous on [a, b].

Note that H (x) = F (g(x)) is continuous on [a, b]. To see this, first note thatg(x) is continuous on [a, b]. Next, note that by the extreme value theorem, f (x)is bounded on g([a, b]), whence corollary 1.4.2 shows that F (t) is (uniformly)continuous on g([a, b]). The composition of F (t) with g(x) will therefore becontinuous on [a, b].

All conditions of the second fundamental theorem of calculus are now satis-fied. By that theorem,

H (b) − H (a) = ba

f (g(x))g(x) dx;

however, we also have

H (b) − H (a) := g(b)g(a)

f (u) du − g(a)g(a)

f (u) du

=

g(b)g(a)

f (u) du.

We conclude that ba

f (g(x))g(x) dx = g(b)g(a)

f (u) du.

Example 1.16. Evaluate e2e 1x ln(x) dx.Let g(x) = ln(x). Then g (x) = 1x ; the change of variables theorem and the

second fundamental theorem of calculus show that e2e

1

x ln(x) dx =

g(e2)g(e)

1

u du

=

ln(e2)ln(e)

1

u du

=

21

1

u du

= ln(

|u

|)

|21

= ln(2) − ln(1)= ln(2).

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1.6 Integration by Parts

Recall that the product rule states that

d

dx f · g =

df

dx

· g +

dg

dx

· f

= f (x)g(x) + g(x)f (x).

Hence for differentiable functions f (x) and g (x), we have that

f (x)g(x) + C =

d

dx (f (x)g(x)) dx =

f (x)g(x) dx +

g(x)f (x) dx,

and thus we get the integration by parts formula

g(x)f (x) dx = f (x)g(x) − f (x)g(x) dx.

(Or equivalently, we can write it this way: f (x)g(x) dx = f (x)g(x) −

g(x)f (x) dx.)

The constant C is accounted for by the indefinite integration sign.Just as integration by substitution was the reversal of the chain rule, this

technique is the reversal of the product rule.


xex dx.

Strategy: We want to rid the integrand of the x. Integration by parts is ideal for

this maneuver. Our way of removing the x is to differentiate it “downwards” toproduce the constant 1. As a compensation, we must integrate the other factorex “upwards”; however, integrating ex is not difficult at all. The integration byparts formula tells us that

xex dx = xex −

1ex dx = xex − ex + C = (x − 1)ex + C.

Sample Solution 2. Let f (x) = x, g (x) = ex. Then f (x) = 1 and g (x) = ex.Integrating by parts, we have

xex dx =

f (x)g(x) dx = f (x)g(x) −

f (x)g(x) dx

= xex − 1ex dx= xex − ex + C = (x − 1)ex + C.

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x2 sin(x) dx.

Strategy: As before, the integrand is a product of two factors, one of whichis easy to remove by bringing it “downwards” by differentiation. On the otherhand, we must bring the other factor, sin(x), “upwards” by integration. Inte-grating sin(x) is not difficult. Once we are done, we end up with this integral:

2x(− cos(x)) dx.

There are still two factors in the integrand. It looks like we need to apply thesame technique one more time: We differentiate 2x “downwards” and integrate− cos(x) “upwards” again, finally eliminating the original factor x2. We get

2(− sin(x)) dx.This integral is easy to evaluate.

Sample Solution 3. Let f 1(x) = x2 and g1(x) = − cos(x). Then f 1(x) = 2x

and g 1(x) = sin(x). Integrating by parts, we get x2 sin(x) dx =

f 1(x)g

1(x) dx = f 1(x)g1(x) −

f 1(x)g1(x) dx

= x2(− cos(x)) −

2x(− cos(x)) dx.

Let f 2(x) = 2x and g2(x) = − sin(x). Then f 2(x) = 2 and g2(x) = − cos(x).Integrating by parts again, we get

2x(− cos(x)) dx =

f 2(x)g2(x) dx = f 2(x)g2(x) −

f 2(x)g2(x) dx

= 2x(− sin(x)) − 2 − sin(x) dx

= −2x sin(x) − 2 cos(x) + C 0.

Putting these together, we have x2 sin(x) dx = x2(− cos(x)) −

2x(− cos(x)) dx

= −x2 cos(x) − [−2x sin(x) − 2cos(x) + C 0]= −x2 cos(x) + 2x sin(x) + 2 cos(x) + C

(where C = −C 0).


ex sin(x) dx.

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Strategy: It does not appear that we can eliminate any of the two factorsthrough integration or differentiation, because ex and sin(x) never vanish after

some differentiation or integration; however, we can use this fact to our advan-tage. Note that ex is “invariant” under integration and sin(x) is “cyclic” underdifferentiation. That is, ex does not change when integrated, and sin(x) first be-comes cos(x) and then becomes − sin(x) when differentiated. To demonstrate,after the first integration by parts, our integral transforms into

ex cos(x) dx.

After the second, we get ex(− sin(x)) dx = −

ex sin(x);

we got back the negative of our original integral. This means we can write theintegral in question in terms of itself and the other items generated along theway, whence we do some algebra.

Observe the sample solution below.

Sample Solution 4. Let f 1(x) = sin(x) and g(x) = ex. Then f 1(x) = cos(x)

and g (x) = ex. Integrating by parts, we get ex sin(x) dx =

g(x)f 1(x) dx = g(x)f 1(x) −

g(x)f 1(x) dx

= ex sin(x) −

ex cos(x) dx.

Let f 2(x) = cos(x). Then f 2(x) = − sin(x). Integrating by parts again, we get

ex cos(x) dx =

g(x)f 2(x) dx = ex cos(x) −

g(x)f 2(x) dx

= ex cos(x) −

ex(− sin(x)) dx

= ex cos(x) +

ex sin(x) dx.

Putting these together, we get

ex sin(x) dx = ex sin(x) − ex cos(x) dx

= ex sin(x) − [ex cos(x) +

ex sin(x) dx]

= ex sin(x) − ex cos(x) −

ex sin(x) dx.

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Moving

ex sin(x) dx to the left side, we obtain

ex sin(x) dx +

ex sin(x) dx = ex sin(x) − ex cos(x),

which implies

2 ·

ex sin(x) dx = ex sin(x) − ex cos(x) + C 0.

We conclude that ex sin(x) dx =

ex sin(x) − ex cos(x)2

+ C

(where C = C 02 ).


ln(x) dx (valid on (0, ∞)).There is an invisible second factor in the integrand: 1 · ln(x). With this

insight, we let f (x) = x and g(x) = ln(x). Then f (x) = 1 and g(x) = 1x .Integrating by parts, we get

ln(x) dx =

f (x)g(x) dx = f (x)g(x) −

g(x)f (x) dx

= x ln(x) −

1

x · x dx

= x ln(x) −

1 dx

= x ln(x) − x + C.Checking, we have that

d

dx [x ln(x) − x + C ] = 1 · ln(x) + x · 1

x − 1 + 0 = ln(x) + 1 − 1 = ln(x).

This indefinite integral is valid on (0, ∞), as required.


arctan(x) dx.Using the same technique, we try to eliminate arctan(x) by differentiation,

while at the same time integrating the invisible factor 1. Let f (x) = x andg(x) = arctan(x). Then f (x) = 1 and g(x) = 11+x2 . Integrating by parts, we

obtain arctan(x) dx =

f (x)g(x) dx = f (x)g(x) −

f (x)g(x) dx

= x arctan(x) −

x

1 + x2 dx.

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Making the substitution u = 1 + x2, we find du = 2x dx and so we now have

x1 − x2 dx = 12

2x1 − x2 dx

= 1

2

1

u du

= 1

2 ln(|u|) + C 0

= 1

2 ln(|1 + x2|) + C 0.

Putting these together, we have that arctan(x) dx = x arctan(x) −

x

1 + x2 dx

= x arctan(x) − 12

ln(|1 + x2|) + C 0= x arctan(x) − 1

2 ln(|1 + x2|) + C

(where C = −C 0).Below are some indefinite integrals that integration by parts will solve par-

ticularly well:

• arcsin(x) dx,•

arccos(x) dx,

• xn sin(x) dx,• xn cos(x) dx,• xnex dx,• cos(x)ex dx,• sin(x)ex dx.

1.7 Partial Fractions

In this section, we will look at how we would integrate rational functions—

functions of the form f (x) = p(x)

q(x)

, where p(x) and q (x)= 0 are both polynomi-

als.

Problem. Evaluate 1x2−1 dx.

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Solution. Note that

1x2 − 1 dx =

1(x + 1)(x − 1) dx,

and so if there exists some A, B ∈ R such that 1(x+1)(x−1) = Ax+1 + Bx−1 , thenwe would have

1

x2 − 1 dx =

1

(x + 1)(x − 1) dx

=

A

x + 1 +

B

x − 1

dx

=

A

x + 1 dx +

B

x − 1 dx= A ln(

|x + 1

|) + B ln(

|x

−1|) + C.

Do such A and B exist? To find them, we write

(1.7) 1

(x + 1)(x − 1) = A

x + 1 +

B

x − 1 = A(x − 1) + B(x + 1)

(x + 1)(x − 1) ,

whence1 = A(x − 1) + B(x + 1).

Setting x = 1, we obtain 1 = 2B, which means B = 12 . Setting x = −1, weobtain 1 = −2A, which means A = −12 . Note that this method is not yet

justified yet, since we would be dividing by 0 if we set x = 1 or x =

−1 at

equation (1.7); however, we can justify this procedure in many ways, and weleave the details as an exercise. Of course, we can always just check that the Aand B does indeed satisfy (1.7):

−12

x + 1 +12

x − 1 =−12 (x − 1) + 12(x + 1)

(x + 1)(x − 1) =−12 x + 12 + 12x + 12

(x + 1)(x − 1) = 1

x2 − 1 .

The solution above therefore yields 1

x2 − 1 dx = A ln(|x + 1|) + B ln(|x − 1|) + C

= −1

2 ln(|x + 1|) + 1

2 ln(|x − 1|) + C.

So the general problem is the following.

Problem. Evaluate p(x)

q(x) dx, where p(x) and q (x) are polynomials (q (x) is not

the zero polynomial).

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Definition 1.7.1. Given a polynomial p(x) = a0 + a1x + a2x2 + · · · + anxn,

where n

≥1 and an

= 0, n is called the degree of p(x) (notation: deg( p(x)) = n).

For p(x) = 0, we define deg( p(x)) := −∞.To simplify the problem, we make the following assumptions and observa-

tions.

1. Without loss of generality, we will assume that deg( p(x)) < deg(q (x)), forotherwise we could use polynomial division to write

p(x) = p1(x) + p2(x)

q (x) ,

where deg( p2(x)) < deg(q (x)). Integrating p1(x) is simple, and so theproblem reduces to the case where deg( p(x)) < deg(q (x)).

2. By the fundamental theorem of algebra , given any polynomial q (x), we canwrite

q (x) = a(q 1(x))m1(q 2(x))

m2 · · · (q k(x))mk ,where each q i(x) = x − ai for some ai ∈ R or q i(x) = x2 + bix + ci forsome bi, ci ∈ R, where q i(x) is irreducible. We will assume without lossof generality that a = 1, so that q (x) is monic. Also, p(x) and q (x) shareno common factors.

1.7.1 Type I Rational Functions

Definition 1.7.2. A rational function f (x) = p(x)q(x) is said to be type I if p(x)

and q (x) are polynomials satisfying

1. deg( p(x)) < deg(q (x)),

2. p(x) = 0 implies q (x) = 0, and3. q (x) = (x − a1)(x − a2) · · · (x − ak) with ai = aj if i = j .

Example 1.19. f (x) = 1x2−1 = 1(x+1)(x−1) is type I.

Remark 1.7.3. If f (x) = p(x)q(x) is type I, then there exists constants A1, A2, . . .,

Ak such that

(1.8) f (x) = A1

x − a1 + A2

x − a2 + · · · + Ak

x − ak ,

where q (x) = (x − a1)(x − a2) · · · (x − ak). The quantity on the right side of (1.8) is called the partial fraction decomposition of f (x).

Proof. This proof is omitted.

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With the constants A1, A2, . . ., Ak described above, we have

f (x) dx =

A1x − a1 +

A2x − a2 + · · · +

Akx − ak

dx

=

A1

x − a1 dx +

A2

x − a2 dx + · · · +

Ak

x − ak dx= A1 ln(|x − a1|) + A2 ln(|x − a2|) + · · · + Ak ln(|x − ak|) + C.

(1.9)

To find A1, A2, · · · , Ak, we multiply both sides of (1.8) by

q (x) = (x − a1)(x − a2) · · · (x − ak)

to get

p(x) =

ki=1

Ai kj=1j=i

(x − aj) .

Setting x = am for m = 1, 2, . . . , k yields

p(am) = Am

kj=1j=m

(am − aj).

Note thatk

j=1j=m

(am − aj) = 0;

therefore, we conclude that

Am = p(am)

C m

where

C m =k

j=1j=m

(am − aj),

for each m = 1, 2, . . . , k.


x(x−1)(x−3) dx.

We write

x

(x − 1)(x − 3) = A

x − 1 + B

x − 3⇒ x = A(x − 3) + B(x − 1).

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Setting x = 1, we get that 1 = A(−2), which means A = −12 . Setting x = 3, weget that 3 = B (2), which means B = 32 . Hence

x

(x − 1)(x − 3) dx = −12

x − 1 dx + 32

x−3x − 3 dx

= −1

2 ln(|x − 1|) + 3

2 ln(|x − 3|) + C.

1.7.2 Type II Rational Functions

Definition 1.7.4. A rational function f (x) = p(x)q(x) is said to be type II if p(x)

and q (x) are polynomials satisfying

1. deg( p(x)) < deg(q (x)),

2. p(x) = 0 implies q (x) = 0,3. q (x) = (x − a1)m1(x − a2)m2 · · · (x − ak)mk with ai = aj if i = j ,4. mi ≥ 1 for all i = 1, 2, . . . k, and5. mi > 1 for some i = 1, 2, . . . k.

In this case, one can show that f (x) = p(x)q(x) has a partial fraction decompo-

sition where each term (x − aj)mj of q (x) contributes mj terms of the formAj,1

x − aj+

Aj,2

(x − aj)2

+ Aj,3

(x − aj)3

+

· · ·+

Aj,mj

(x − aj)mj

to the decomposition.Integrating this type of rational functions is very similar to integrating those

of type I. We give an example below.

Example 1.21. Evaluate 1x2(x+1)

dx.

Here, q (x) = x2(x + 1), making 1x2(x+1) a type II rational function; the x2

will contribute two terms to the decomposition while the (x + 1) will contributeone term:

1

x2(x + 1) =

A1,1

x +

A1,2

x2 +

A2,1

x + 1

:= A

x +

B

x2 +

C

x + 1.(1.10)

To find A, B , and C , we multiply the entire equation by q (x) = x2(x +1) to get

(1.11) 1 = Ax(x + 1) + B(x + 1) + Cx2.

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Letting x = 0, we get that 1 = 0 + B · 1 + 0 = B . Letting x = −1, we obtain1 = 0 + 0 + C

·1 = C . Finally, we find A by comparing coefficients in equation

(1.11):

1 = Ax2 + Ax + Bx + B + Cx2 = (A + 1)x2 + (A + 1)x + 1

⇒ 0 = (A + 1)x2 + (A + 1)x⇒ 0x2 = (A + 1)x2⇒ 0 = A + 1⇒ A = −1.

After checking, we find that this combination of A, B, and C satisfies (1.10).We therefore have that

1

x2

(x + 1)

dx =

− 1

x

dx + 1

x2

dx + 1

x + 1

dx

= − ln(|x|) − 1x

+ ln(|x + 1|) + C.

1.7.3 Type III Rational Functions

Definition 1.7.5. A rational function f (x) = p(x)q(x) is said to be type III if p(x)

and q (x) are polynomials and f (x) is neither type I nor type II. That is, f (x)satisfies

1. deg( p(x)) < deg(q (x)),

2. p(x) = 0 implies q (x) = 0,3. q (x) is of the form

q (x) =

(x2 + b1x + c1)m1(x2 + b2x + c2)

m2 · · · (x2 + blx + cl)ml×

[(x − a1)ml+1(x − a2)ml+2 · · · (x − ak)ml+k ]

with 0 < l ≤ k and mi > 0 for all i = 1, 2, . . . , l + k,4. ai = aj if i = j , for l ≤ i, j ≤ k,5. (bi, ci) = (bj , cj) if i = j , for 1 ≤ i, j ≤ l,6. (x2 + bix + ci) is irreducible for all i ∈ {1, 2, . . . , l}, and7. (x2 + bix + ci) is not a polynomial factor of p(x) for any i ∈ {1, 2, . . . , l}.Essentially, type III rational functions have at least one irreducible quadratic

factor in their denominators. One can show that each term (x2+ bjx + cj)mj of

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q (x) will contribute mj terms to the partial fraction decomposition of f (x) asfollows:

Aj,1x + Bj,1x2 + bjx + cj

+ Aj,2x + Bj,2(x2 + bjx + cj)2

+ Aj,3x + Bj,3(x2 + bjx + cj)3

+ · · · + Aj,mjx + Bj,mj(x2 + bjx + cj)mj

.

The linear factors of q (x) contribute terms to the decomposition in the samemanner as before.

Example 1.22. Suppose that f (x) = p(x)q(x) is a type III rational function where

q (x) = x(x2 + 1)2. Its partial decomposition takes the form

p(x)

q (x) =

A1

x +

A2x + B2x2 + 1

+ A3x + B3(x2 + 1)2

.

Once we find the constants in the decomposition, we begin integrating. Thelinear terms are integrated as before; the quadratic terms are handled by thefollowing simple fact.

Example 1.23. 1x2+1 = arctan(x) + C .

¿From this, we can solve the following.

Example 1.24. Evaluate 1x2+x+1 dx.

Note that this type III rational function is already in its decomposition. We

first complete the square: x2+ x + 1 = x + 122+ 34 . Then, we factor the 34 outto get

x2 + x + 1 = 3

4

4

3

x +

1

2

2+ 1

,

and so we get1

x2 + x + 1 =

4

3 · 1 2√ 3

x + 12

2+ 1

.

Let u = 2√ 3

x + 12

, so that du = 2√

3 dx ⇒ dx =

√ 32

du. Integrating, we get

1x2 + x + 1 dx = 4

3 · 1 2√ 3 x + 122 + 1 dx

= 4

3 ·

1

u2 + 1 ·

√ 3

2 du

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= 4

3 ·

√ 3

2 1

u2 du

= 2√

3· (arctan(u) + C 0)

= 2√

3· arctan

2√

3

x +

1

2

+ C.

In general, we get the following.

Example 1.25. Evaluate 1

x2+bx+c dx, where b2 − 4c < 0 (i.e., x2 + bx + c is

irreducible).Note that we can do the same algebra as the previous example to obtain

1

x2 + bx + c =

1

D

1 1√ D

x + b2

2+ 1

,where D = c − b24 > 0. Making the substitution u =

1√ D

x + b2

, we get that

du = 1√ D

dx ⇒ dx = √ D du, and so 1

x2 + bx + c dx =

1

D ·

1 1√ D

x + b2

2+ 1

dx

= 1

D ·

1

u2 + 1 ·

√ D du

=√ D

D ·

1

u2 + 1 du

= 1√

D(arctan(u) + C 0)

= 1√

Darctan

1√

D

x +

b

2

+ C.

We have found the general formula 1

x2 + bx + c dx =

1√ D

arctan

1√

D

x +

b

2

+ C,

where D = c − b2

4 > 0.We now look at how we would integrate an irreducible quadratic term with

a linear polynomials as its numerator.


xx2+1 dx.

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Let u = x2 + 1; then du = 2x dx ⇒ x dx = du2 . Integrating, we have x

x2 + 1 dx =

1u · du

2

= 1

2

1

u du

= 1

2(ln(|u|) + C 0)

= 1

2 ln(|x2 + 1|) + C.


x+1x2+1 dx.

We first break this up into a sum of two integrals: x + 1

x2 + 1 dx =

x

x2 + 1 dx +

1

x2 + 1 dx,

both of which we have already integrated in previous examples. We get that x + 1

x2 + 1 dx =

1

2 ln(|x2 + 1|) + arctan(x) + C.


xx2+x+1 dx.

We again break this up into two integrals, keeping in mind that the substi-tution u = x2 + x + 1 has differential du = (2x + 1) dx

⇒ du

2 = x + 12 dx. Wehave

x

x2 + x + 1 dx =

x + 12x2 + x + 1

dx +

−12x2 + x + 1

dx

=

1

u · du

2 − 1

2 ·

1

x2 + x + 1 dx

= 1

2 ln(|u|) − 1√

3· arctan

2√

3

x +

1

2

+ C

= 1

2 ln(|x2 + x + 1|) − 1√

3arctan

2√

3

x +

1

2

+ C.

We now have an algorithm to integrate every irreducible quadratic term of the type Ax+Bx2+bx+c . But what about the terms of the form

Ax+B(x2+bx+c)n , where

n > 1? We can break this up into two integrals Ax + Ab2(x2 + bx + c)n

dx +

b − Ab

2

·

1

(x2 + bx + c)n dx.

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Here, we deliberately made the first integral easy to integrate: simply let u =x2 + bx + c and we get du = (2x + b) dx

⇒ A2 du = Ax + Ab2 dx. The integralon the left now becomes the easy integral

1

un · A

2 du =

A

2 ·

1

un du.

The integral on the right is the more difficult one, and it would require techniquesfrom the next section to solve.

1.8 Inverse Trignometric Substitutions

Problem. Evaluate 1(x2+1)2

dx.

Solution. Here, we make the substitution θ = arctan(x) ⇒ x = tan(θ). Weget that dx = sec2(θ) dθ and as a result

1

(x2 + 1)2 dx =

sec2(θ)

(tan2(θ) + 1)2 dθ

=

sec2(θ)

(sec2(θ))2 dθ

=

1

sec2(θ) dθ

=

cos2(θ) dθ

= cos(2θ)2 + 12 dθ=

1

2 · sin(2θ)

2 +

θ

2 + C

= 1

4(sin(2θ)) +

θ

2 + C,

since cos2(θ) = cos(2θ)+12 . Recall that x = tan(θ); from figure 1.1, we can see

1

1+x 2

x

θ

Figure 1.1: The subtitution θ = arctan(x).

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that sin(θ) = x√ x2+1

and cos(θ) = 1√ x2+1

. Then sin(2θ) = 2sin(θ) cos(θ) =2x

x2

+1. We therefore have

1

(x2 + 1)2 dx =

1

4(sin(2θ)) +

θ

2 + C =

x

2(x2 + 1) +

1

2 arctan(x) + C.

The substitution θ = arctan(x) is called an inverse trigonometric substitu-tion , or simply an inverse trig substitution. If we rewrite this particular substitu-tion as x = tan(θ) and then take advantage of the identity sec2(θ) = 1+ tan2(θ),we can evaluate many similar integrals.

Note. If the integrand has terms of the form (a2x2 + b2)α, then make thesubstitution θ = arctanaxb ⇒ ax = b tan(θ). Use the substitution ax =b tan(θ) to obtain

(a2x2 + b2)α = (b2 tan2(θ) + b2)α = b2α(tan2(θ) + 1)α = b2α(sec2(θ))α.

Example 1.29. One should make the θ = arctan x substitution to evaluate 1√ x2+1

dx.

Problem. Evaluate the definite integral

1

−1√

1 − x2 dx.

Solution. Here, we make the substitution θ = arcsin(x) ⇒ x = sin(θ). Thendx = cos(θ) dθ, valid on

−π2

, −π2

. Note that cos(θ) ≥ 0 on that interval. We

now have, by the change of variables theorem, that 1−1

1 − x2 dx =

arcsin(1)arcsin(−1)

1 − sin2(θ) · cos(θ) dθ

=

π2

−π2

cos2(θ) · cos(θ) dθ

=

π2

−π2

| cos(θ)| cos(θ) dθ

= π2

−π2

cos(θ) · cos(theta) dθ

=

π2

−π2

cos2(θ) dθ.

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We now use the fact that cos2(θ) = cos(2θ)+12 = cos(2θ)2 + 12 to get

π2−π2

cos2(θ) dθ = 12 · π2−π2

cos(2θ) dθ + π2−π2

12

dθ

= sin(2θ)

4

π2

−π2

+ θ

2

π2

−π2

=

π

4 − −π

4

+

sin

2 · π2

4

− sin

2 · −π2

4

= π

2 + (0 − 0)

= π

2.

We get that the area of the semicircle of radius 1 is π

2 .The above solution demonstrates another inverse trig substitution: θ =

arcsin(x). This substitution is used so that we could take advantage of theidentity 1 − sin2(x) = cos2(x).

Note. If the integrand has terms of the form (b2 − a2x2)α, then make thesubstitution θ = arcsin

axb

⇒ ax = b sin(θ). Use the substitution ax = b sin(θ)to obtain

(b2 − a2x2)α = (b2 − b2 sin2(θ))α = b2α(1 − sin2(θ))α = b2α(cos2(θ))α.

Problem. Evaluate 1√ x2−1 dx.

Solution. Here, we make the substitution θ = arcsec(x) ⇒ x = sec(θ). Thendx = sec(θ) tan(θ) dθ, valid on

0, π2

. Now tan(θ) ≥ 0 on that interval; hence,

we have that 1√

x2 − 1 dx =

sec(θ) tan(θ) sec2(θ) − 1 dθ

=

sec(θ) tan(θ)√

tan2 θdθ

= sec(θ) tan(θ)

tan θ dθ

=

sec(θ) dθ

= ln(| sec(θ) + tan(θ)|) + C.

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Recall that x = sec(θ); from the fact that tan(θ) =√

sec2 θ − 1 = √ x2 − 1, weget

1√ x2 − 1 dx = ln(| sec(θ) + tan(θ)|) + C = ln

x + x2 − 1 + C.

The above solution demonstrates the use of the substitution θ = arcsec(x).It is most often used in conjunction with the identity sec2(x) − 1 = tan2(x).

Note. If the integrand has terms of the form (a2x2 − b2)α, then make thesubstitution θ = arcsec

axb

⇒ ax = b sec(θ). Use the substitution ax = b sec(θ)to obtain

(a2

x2

− b2

)α

= (b2

sec2

(θ) − b2

)α

= b2α

(sec2

(θ) − 1)α

= b2α

(tan2

(θ))α

.

With these techniques, we can now integrate any rational function f (x) = p(x)q(x) (where q (x) is not the zero polynomial).

1.9 Applications of the Integral

1.9.1 Area Between Curves

Problem. Let R be the region bounded by the graphs of f (x) and g(x) andthe lines x = a and x = b. Suppose that f (x) and g(x) are continuous on [a, b].

Find the area A of region R (figure 1.2).

g x( )

f x( )

x

y

a bR

Figure 1.2: Area between curves.

For convenience, we will in the future say that the region R is bounded by (the graphs of) f (x) and g(x) on [a, b].

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Solution. We find A in three steps.

1. We begin by partitioning the interval [a, b] using

P := {xi : a = x0 < x1 < · · · < xi−1 < xi < · · · < xn = b}.Let Ri be the subregion of R bounded by the graphs of y = g(x) y = f (x)

g x( )

f x( )

x

y

a bxi-1 xi

Ri

Figure 1.3: Partitioning R into subregions.

on [xi−1, xi]. Let Ai be the area of Ri. Then A = n

i=1Ai. Figure 1.3shows this step.

2. We estimate each Ai with the area of a rectangle with its base on [xi−1, xi]and its top and bottom on the lines y = max{g(xi), f (xi)} and y =min{g(xi), f (xi)}, respectively. This rectangle has width ∆xi := xi−xi−1and height

|g(xi)

−f (xi)

|. Thus,

Ai ≈ |g(xi) − f (xi)|∆xi,so that

A =n

i=1

Ai ≈n

i=1

|g(xi) − f (xi)|∆xi.

3. If f (x) and g(x) are continuous on [a, b] and if P n is the n-regular partition,then letting n → ∞ gives

A = limn→∞

ni=1

Ai

= limn→∞

ni=1

ga + i b − an − f a + i b − an · b − an=

ba

|g(x) − f (x)| dx,

where the last equality is given by corollary 1.2.10.

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The above solution can be used as the definition of the area bounded by (the graphs of) f (x) and g(x) on [a, b], where f (x) and g(x) are continuous functions.

If f (x) and g(x) intersect at least twice and at most a finite number of times,then we can define the area bounded by (the graphs of) f (x) and g(x) as thearea bounded by f (x) and g(x) on [c, d], where x = c is the leftmost point wherethe two functions intersect and d is the rightmost point where the two functionsintersect.

Example 1.30. Let f (x) = x, g(x) = x3. Find the area bounded by f (x) andg(x).

Since we are not given bounds a and b, we are finding the total boundedarea. We note that f (x) = x and g(x) = x3 intersect three times: at x = −1,at x = 0, and at x = 1. To see this, note that x = x3 implies that x3− x = 0 ⇒x(x + 1)(x

−1) = 0. Let A be the area bounded by the two graphs. Then we

have that

A =

1−1

x3 − x dx=

0−1

x3 − x dx + 10

x3 − x dx=

0−1

(x3 − x) dx + 10

(x − x3) dx

=

x4

4 − x2

2

0−1

+

x2

2 − x4

4

10

(1.12)

= 1

4

+ 1

4=

1

2.

Equality (1.12) is due to the second fundamental theorem of calculus.

1.9.2 Moments and Centers of Mass

Informal definition. A moment is the product of the mass of an object timesthe “directed” distance from a fixed point.

Assume that R is a region in R2 bounded by f (x) and g(x) on [a, b] withf (x) ≤ g(x) on [a, b]. Assume that f (x) and g (x) are continuous on [a, b]. Alsoassume that the region R has area A and represents a thin plate of uniformdensity ρ and uniform thickness t. Then the mass of the plate, m, is given by

m = ρtA.

Without loss of generality, we may assume that the constant ρt is equal to 1, sothat

m = A =

ba

(g(x) − f (x)) dx.

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Problem. Find the moment of R relative to the y -axis.

Solution. As before, we first partition the interval [a, b] into n subintervalsusing

P := {xi : a = x0 < x1 < · · · < xi−1 < xi < · · · < xn = b}This subdivides R into n regions Ri, i = 1, 2, . . . , n. The moment M i associatedwith the region Ri is approximately

M i ≈ mici = ci(g(ci) − f (ci))∆xi,

where ci = xi−1+xi2 represents the “x center,” or the “average distance from the

y-axis,” of the rectangle with its base on [xi−1, xi] and its top and bottom atg(ci) and f (ci), respectively; mi = (g(ci) − f (ci))∆xi is the area, which is themass, of this rectangle. The total moment of R relative to the y-axis , M x, is the

sum of the individual moments

M x =n

i=1

M i ≈n

i=1

ci(g(ci) − f (ci))∆xi.

(See figure 1.4.) Letting n → ∞ gives

g x( )

f x( )

x

y

a bxi-1 xi

g( )ci

f ( )ci

ci

Figure 1.4: ci gives the shaded rectangle’s “average distance” from the y -axis.

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M x = limn

→∞

n

i=1M i

= limn→∞

ni=1

ci(g(ci) − f (ci))∆xi

=

ba

x(g(x) − f (x)) dx.

The last equality is due to corollary 1.2.10. If we remove the assumption thatf (x) ≤ g(x), then we can see that we would get

M x =

ba

x|g(x) − f (x)| dx.

Informal definition. Suppose R is a region in R2 bounded by f (x) and g(x)on [a, b], where f (x) and g(x)

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