Chapter 5: Integration

Fall 2014

Department of MathematicsHong Kong Baptist University

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§5.1 Sums and Sigma Notation

Definition (Sigma Notation)

If m and n are integers with m ≤ n, and if f is a function definedat the integers m,m + 1, . . . , n, the symbol

∑ni=m f (i) represents

the sum of the values of f at those integers:

n∑i=m

f (i) = f (m) + f (m + 1) + · · ·+ f (n).

The explicit sum appearing on the right side of the equation is theexpansion of the sum represented in sigma notation on the leftside. m is the lower limit, and n is the upper limit.

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Theorem (Summation Formulas)

(a)n∑

i=1

i = 1 + 2 + · · ·+ n =n(n + 1)

2.

(b)n∑

i=m

i = m + (m + 1) + · · ·+ n =(n −m + 1)(n + m)

2.

(c)n∑

i=1

i2 = 12 + 22 + · · ·+ n2 =n(n + 1)(2n + 1)

6.

(d)n∑

i=1

i3 = 13 + 23 + · · ·+ n3 =n2(n + 1)2

4.

(e)n∑

i=1

r i−1 = 1 + r + r2 + · · ·+ rn−1 =rn − 1

r − 1.

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§5.2 Areas as Limits of Sums

This section aims to find the area of a region R lying under thegraph y = f (x) of a nonnegative-valued, continuous function f ,above the x-axis and between the vertical lines x = a and x = b.

- x

6y

y = f (x)

a b

R

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To compute the area, we divide the intervals [a, b] into nsubintervals using division points:

a = x0 < x1 < x2 < · · · < xn−1 < xn = b.

Denote by ∆xi the length of the ith subinterval [xi−1, xi ]:

∆xi = xi − xi−1, i = 1, . . . , n.

The area of the ith rectangle is f (xi )∆xi . The sum of all theseareas is

Sn =n∑

i=1

f (xi )∆xi .

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It is often convenient to choose the points xi so that ∆xi areall equal. In this case we have

∆xi = ∆x =b − a

nor xi = a +

i

n(b − a).

Sn is an approximation to the area of the region R, and theapproximation gets better as n increases. That is,

Area of R = limn→∞

Sn

= limn→∞

b − a

n

n∑i=1

f (xi ).

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Example 1:

Find the area of the region bounded by the parabola y = x2 andthe straight lines y = 0, x = 0, and x = b, where b > 0.

Solution: Let f (x) = x2. We use equal subintervals, each oflength b/n and xi = ib/n. Thus,

Sn =b − 0

n

n∑i=1

f (xi ) =b3

n3

n∑i=1

i2 =b3(n + 1)(2n + 1)

6n2.

Therefore, the required area A is

A = limn→∞

Sn

= limn→∞

b3(n + 1)(2n + 1)

6n2

=b3

3.

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§5.3 The Definite Integral

In this section we define the definite integral of a continuousfunction f on an closed, finite interval [a, b]. We no longerassume that the values of f are nonnegative.

Let P be a finite set of points arranged in order between aand b on the real line, say P = {x0, x1, x2, . . . , xn−1, xn},where a = x0 < x1 < x2 < · · · < xn−1 < xn = b.

Such a set P is called a partition of [a, b] and it divides [a, b]into n = n(P) subintervals. The ith subinterval of P is[xi−1, xi ] and its length is

∆xi = xi − xi−1, 1 ≤ i ≤ n.

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We call the greatest of these numbers ∆xi the norm of thepartition P and denote it by

‖P‖ = max1≤i≤n

∆xi .

Since f is continuous on each [xi−1, xi ], there are numbers liand ui in [xi−1, xi ] such that

f (li ) ≤ f (x) ≤ f (ui ), xi−1 ≤ x ≤ xi .

If f (x) ≥ 0 on [a, b] and if Ai is the part of the area boundedby y = f (x), y = 0, x = xi−1 and x = xi , then

f (li )∆xi ≤ Ai ≤ f (ui )∆xi .

If f can take negative values, f (li )∆xi ≤ f (ui )∆xi still holds.

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Upper and Lower Riemann Sums

Definition

The lower (Riemann) sum, L(f ,P), and the upper (Riemann)sum, U(f ,P), for the function f and the partition P are definedby:

L(f ,P) = f (l1)∆x1 + f (l2)∆x2 + · · ·+ f (ln)∆xn

=n∑

i=1

f (li )∆xi ,

and

U(f ,P) = f (u1)∆x1 + f (u2)∆x2 + · · ·+ f (un)∆xn

=n∑

i=1

f (ui )∆xi .

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Example 2:

Calculate the lower and upper Riemann sums for the functionf (x) = x2 on the interval [0, b] where b > 0, corresponding to thepartition Pn of [0, b] into n subintervals of equal length.

Solution: With equal length subintervals, we have ∆x = b/n andthe division points are xi = ib/n for i = 0, 1, . . . , n.

Since f (x) = x2 is increasing on [0, b], its minimum and maximumvalues over the ith subintervals [xi−1, xi ] occur at li = xi−1 andui = xi , respectively. Thus, the lower Riemann sum of f for Pn is

L(f ,Pn) =n∑

i=1

(xi−1)2∆x =(n − 1)(2n − 1)b3

6n2,

and the upper Riemann sum is

U(f ,Pn) =n∑

i=1

(xi )2∆x =

(n + 1)(2n + 1)b3

6n2.

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If P1 and P2 are two partitions of [a, b] such that every pointsof P1 also belongs to P2, i.e. P1 ⊂ P2, then we say that P2 isa refinement of P1. It is not difficult to show that

L(f ,P1) ≤ L(f ,P2) ≤ U(f ,P2) ≤ U(f ,P1);

that is, adding more points to a partition increases the lowersum and decreases the upper sum.

If P is partitioned to have more and more points spaced closerand closer together, in the limit sense we expect that

L(f ,Pn) ≤ limn→∞

L(f ,Pn) ≤ limn→∞

U(f ,Pn) ≤ U(f ,Pn).

If limn→∞

L(f ,Pn) = limn→∞

U(f ,Pn) , I , we will call I the

definite integral of f on [a, b].

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Definition (The Definite Integral)

Suppose there is exactly one number I such that for every partitionP of [a, b] we have

L(f ,P) ≤ I ≤ U(f ,P).

Then we say that the function f is integrable on [a, b], and wecall I the definite integral of f on [a, b]. The definite integral isdenoted by the symbol

I =

∫ b

af (x)dx .

Remark: The definite integral of f (x) over [a, b] is a number butnot a function of x. Replacing x with another variable does notchange the value of the integral. E.g.,

∫ ba f (x)dx =

∫ ba f (t)dt.

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The various parts of the symbol

∫ b

af (x)dx have their own names:

(i)∫

is called the integral sign, it resembles the letter S since itrepresents the limit of a sum.

(ii) a and b are called the limits of integration; a is the lowerlimit and b is the upper limit.

(iii) The function f is the integrand and x is the variable ofintegration.

(iv) dx is the differential of x . It replaces ∆x in the Riemannsums.

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Example 3:

Show that f (x) = x2 is integrable over the interval [0, b], and

evaluate

∫ b

0f (x)dx .

Solution: By Example 2, we have

limn→∞

L(f ,Pn) = limn→∞

(n − 1)(2n − 1)b3

6n2=

b3

3,

limn→∞

U(f ,Pn) = limn→∞

(n + 1)(2n + 1)b3

6n2=

b3

3.

Then if L(f ,Pn) ≤ I ≤ U(f ,Pn), we must have I = b3/3. Thisshows that f (x) = x2 is integrable over [0, b], and∫ b

0f (x)dx =

∫ b

0x2dx =

b3

3.

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§5.4 Properties of the Definite Integral

Theorem

Let f and g be integrable on an interval containing the points a, b,and c. Then

(a) An integral over an interval of zero length is zero,∫ a

af (x)dx = 0.

(b) Reversing the limits of integration changes the sign of theintegral, ∫ a

bf (x)dx = −

∫ b

af (x)dx .

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Theorem

(c) If A and B are constants, then∫ b

a(Af (x) + Bg(x)) dx = A

∫ b

af (x)dx + B

∫ b

ag(x)dx .

(d) An integral depends additively on the interval of integration,∫ b

af (x)dx +

∫ c

bf (x)dx =

∫ c

af (x)dx .

(e) If a ≤ b and f (x) ≤ g(x) for a ≤ x ≤ b, then∫ b

af (x)dx ≤

∫ b

ag(x)dx .

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Theorem

(f) If a ≤ b, then ∣∣∣∣∫ b

af (x)dx

∣∣∣∣ ≤ ∫ b

a|f (x)|dx .

(g) If f is an odd function, i.e., f (−x) = −f (x), then∫ a

−af (x)dx = 0.

(h) If f is an even function, i.e., f (−x) = f (x), then∫ a

−af (x)dx = 2

∫ a

0f (x)dx .

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Theorem (The Mean-Value Theorem for Integrals)

If f is continuous on [a, b], then there exists a point c in [a, b] suchthat ∫ b

af (x)dx = (b − a)f (c).

Proof: Let f (l) be the minimum value and f (u) be the maximumvalue of f on [a, b]. By the property (e), We have

f (l) =1

b − a

∫ b

a

f (l)dx ≤ 1

b − a

∫ b

a

f (x)dx ≤ 1

b − a

∫ b

a

f (u)dx = f (u).

Then by the Intermediate-Value Theorem, there exists a number cbetween l and u such that

f (c) =1

b − a

∫ b

af (x)dx .

This proves the theorem.23 / 46

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Average Value of a Function

Definition

If f is integrable on [a, b], then the average value or mean valueof f on [a, b], denoted by f̄ , is

f̄ =1

b − a

∫ b

af (x)dx .

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Piecewise Continuous Functions

Example 4:

Find

∫ 3

0f (x)dx , where f (x) =

√

1− x2 if 0 ≤ x ≤ 12 if 1 < x ≤ 2x − 2 if 2 < x ≤ 3.

Solution: By definition, the value of the integral is∫ 3

0f (x)dx =

∫ 1

0

√1− x2dx +

∫ 2

12dx +

∫ 3

2(x − 2)dx

=π

4+ 2 +

1

2

=π + 10

4.

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§5.5 The Fundamental Theorem of Calculus

The following Fundamental Theorem of Calculusdemonstrates the relationship between the definite integraldefined in §5.3 and the indefinite integral (or antiderivative)introduced in §2.10.

Both conclusions of the Fundamental Theorem are useful:

(i) Part I concerns the derivative of an integral; it tells youhow to differentiate a definite integral with respect to itsupper limit.

(ii) Part II concerns the integral of a derivative; it tells youhow to evaluate a definite integral if you can find anantiderivative of the integrand.

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Theorem (The Fundamental Theorem of Calculus)

Suppose that f is continuous on an interval I containing point a.

Part I. Let the function F be defined on I by

F (x) =

∫ x

af (t)dt.

Then F is differentiable on I , and F ′(x) = f (x) there. Thus,F is an antiderivative of f on I :

d

dx

∫ x

af (t)dt = f (x).

Part II. If G is any antiderivative of f on I , so that G ′(x) = f (x) on I ,then for any b in I we have∫ b

af (x)dx = G (b)− G (a).

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Proof: [Part I] Using the definition of the derivative, we have

F ′(x) = limh→0

F (x + h)− F (x)

h

= limh→0

1

h

(∫ x+h

af (t)dt −

∫ x

af (t)dt

)= lim

h→0

1

h

∫ x+h

xf (t)dt.

By the Mean-Value Theorem for Integrals, there exists a pointc ∈ [x , x + h] such that∫ x+h

xf (t)dt = hf (c).

This leads to

F ′(x) = limh→0

f (c) = limc→x

f (c) = f (x).

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[Part II] If G (x) is any antiderivative of f (x) so that G ′(x) = f (x),then (F (x)− G (x))′ = F ′(x)− G ′(x) = 0. This leads to

F (x) = G (x) + C .

Hence, ∫ x

af (t)dt = F (x) = G (x) + C .

Letting x = a, we have 0 = G (a) + C and so C = −G (a). Now welet x = b, then∫ b

af (t)dt = G (b) + C = G (b)− G (a).

Note that t can be replaced by x (or any other variable) as thevariable of integration on the left-hand side. This proves Part II.

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Definition

To facilitate the evaluation of definite integrals using theFundamental Theorem of Calculus, we define the evaluationsymbol:

F (x)∣∣∣ba

= F (b)− F (a).

Remark: By definition, we have∫ b

af (x)dx =

(∫f (x)dx

)∣∣∣∣ba

,

where∫

f (x)dx denotes the indefinite integral or generalantiderivative of f . When evaluating a definite integral this way,we can omit the constant of integration from the indefinite integralsince

(F (x) + C )∣∣∣ba

= F (b) + C − (F (a) + C ) = F (x)∣∣∣ba.

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Example 5:

Find the integral of

∫cosh(x)dx , where cosh(x) =

1

2(ex + e−x).

Solution: We have∫cosh(x)dx =

∫1

2(ex + e−x)dx

=1

2(

∫exdx +

∫e−xdx)

=1

2(ex − e−x) + C

= sinh(x) + C ,

where sinh(x) =1

2(ex − e−x).

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Example 6:

Find the following integrals:

(a)

∫ π

0sin(x)dx = − cos(x)|π

0= 2

(b)

∫ 0

−πsin(x)dx = − cos(x)|0−π

= −2

(c)

∫ π

−πsin(x)dx = − cos(x)|π−π

= 0

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Example 7:

Find the area A of the plane region lying above the x-axis andunder the curve y = 3x − x2.

Solution: Noting that y = 3x − x2 = x(3− x), the two interceptson the x-axis are (0, 0) and (3, 0). The area of the region is thengiven by

A =

∫ 3

0(3x − x2)dx

=

(3

2x2 − 1

3x3

)∣∣∣∣30

=27

2− 27

3− (0− 0)

=9

2.

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Example 8:

Using the conclusion in Part I of the Fundamental Theorem to findthe derivatives of the following functions:

(a) F (x) =

∫ 3

xe−t

2dt, (b) G (x) = x2

∫ x

−4e−t

2dt.

Solution:

(a) Note that F (x) = −∫ x3 e−t

2dt. By Part I we have

F ′(x) = − d

dx

∫ x

3e−t

2dt = −e−x

2.

(b) By the Product Rule and Part I, we have

G ′(x) = 2x

∫ x

−4e−t

2dt + x2 d

dx

∫ x

−4e−t

2dt

= 2x

∫ x

−4e−t

2dt + x2e−x

2.

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By building in the Chain Rule into the conclusion in Part I ofthe Fundamental Theorem, we have the following formula:

d

dx

∫ g(x)

af (t)dt = f (g(x))g ′(x).

More generally, if the lower limit is also a function of x , then

d

dx

∫ g(x)

h(x)f (t)dt = f (g(x))g ′(x)− f (h(x))h′(x).

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Example 9:

Find the derivatives of the following functions:

(a) F (x) =

∫ 5x

−4e−t

2dt, (b) G (x) =

∫ x3

x2e−t

2dt.

Solution:

(a) Let the integrand be f (t) = e−t2

and the upper limit beg(x) = 5x . Then

F ′(x) = f (g(x))g ′(x) = e−(5x)2(5) = 5e−25x

2.

(b) Note that G (x) =∫ x3

0 e−t2dt −

∫ x2

0 e−t2dt. We have

G ′(t) = e−(x3)2(3x2)− e−(x

2)2(2x)

= 3x2e−x6 − 2xe−x

4.

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§5.6 The Method of Substitution

In this section and in Chapter 6 we will develop sometechniques of integration, that is, methods for findingantiderivatives of functions.

Let’s begin by assembling a table of some known indefiniteintegrals. These results have all emerged during ourdevelopment of differentiation formulas for elementaryfunctions and we should memorize them.

Note that formulas 1-6 are special cases of formula 7. Thefollowing linearity formula also makes it possible to integratesums and constant multiples of functions:∫

(Af (x) + Bg(x))dx = A

∫f (x)dx + B

∫g(x)dx .

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When an integral cannot be evaluated by inspection, werequire one or more special techniques. The most importantof these techniques is the method of substitution, theintegral version of the Chain Rule.

If we rewrite the Chain rule,d

dxf (g(x)) = f ′(g(x))g ′(x), in

integral form, we obtain∫f ′(g(x))g ′(x)dx = f (g(x)) + C .

[Another approach to the above formula] Let u = g(x). Thendu/dx = g ′(x), or equivalently, du = g ′(x)dx . Thus∫

f ′(g(x))g ′(x)dx =

∫f ′(u)du = f (u) + C = f (g(x)) + C .

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Example 10:

Find the indefinite integral I =

∫ex√

1 + exdx .

Solution: Let u = 1 + ex . Then du = exdx . Using the method ofsubstitution,

I =

∫ √udu

=2

3u3/2 + C

=2

3(1 + ex)3/2 + C .

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Example 11:

Find the indefinite integral I =

∫1√

e2x − 1dx .

Solution: An appropriate substitution seems not obvious here.Note however that

I =

∫1

ex√

1− e−2xdx =

∫e−x√

1− e−2xdx .

Let u = e−x . Then du = −e−xdx . Using the method ofsubstitution,

I = −∫

1√1− u2

du

= − sin−1(u) + C

= − sin−1(e−x) + C .

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Theorem (Substitution of Definite Integrals)

Suppose that g is a differentiable function on [a, b] that satisfiesg(a) = A and g(b) = B. Also suppose that f is continuous on therange of g. Then∫ b

af (g(x))g ′(x)dx =

∫ B

Af (u)du.

Proof: Let F be an antiderivative of f , that is, F ′(u) = f (u).Then

d

dxF (g(x)) = F ′(g(x))g ′(x) = f (g(x))g ′(x).

Thus,∫ b

af (g(x))g ′(x)dx = F (g(x))

∣∣∣ba

= F (g(b))− F (g(a))

= F (B)− F (A) =

∫ B

Af (u)du.

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Example 12:

Find the integral of

∫ 3

2

x2

x3 − 1dx .

Solution: Let u = x3 − 1. Then

du = 3x2dx .

If x = 2, then u = 7; if x = 3, then u = 26. Using the method ofsubstitution, ∫ 3

2

x2

x3 − 1dx =

1

3

∫ 26

7

1

udu

=1

3ln(u)

∣∣∣∣267

=1

3ln(

26

7).

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Example 13:

Evaluate the integral I =

∫ 8

0

cos√

x + 1√x + 1

dx .

Solution: Let u =√

x + 1. Then

du =1

2√

x + 1dx .

If x = 0, then u = 1; if x = 8, then u = 3. Using the method ofsubstitution,

I = 2

∫ 3

1cos(u)du

= 2 sin(u)∣∣∣31

= 2 (sin(3)− sin(1)) .

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