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Section 5.2The Definite Integral
V63.0121.006/016, Calculus I
New York University
April 15, 2010
Announcements
I April 16: Quiz 4 on §§4.1–4.4
I April 29: Movie Day!!
I April 30: Quiz 5 on §§5.1–5.4
I Monday, May 10, 12:00noon (not 10:00am as previously announced)Final Exam
Announcements
I April 16: Quiz 4 on§§4.1–4.4
I April 29: Movie Day!!
I April 30: Quiz 5 on§§5.1–5.4
I Monday, May 10, 12:00noon(not 10:00am as previouslyannounced) Final Exam
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 2 / 28
Objectives
I Compute the definiteintegral using a limit ofRiemann sums
I Estimate the definiteintegral using a Riemannsum (e.g., Midpoint Rule)
I Reason with the definiteintegral using its elementaryproperties.
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 3 / 28
Notes
Notes
Notes
1
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Outline
Recall
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 4 / 28
Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to findthe area between x = a, x = b, y = 0, and y = f (x).For each positive integer n, divide up the interval into n pieces. Then
∆x =b − a
n. For each i between 1 and n, let xi be the ith step between a
and b. So
xx0 x1 xi xn−1xn. . . . . .
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n. . .
xi = a + i · b − a
n. . .
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 5 / 28
Forming Riemann sums
We have many choices of representative points to approximate the area ineach subinterval.
. . . even random points!
xIn general, choose ci to be a point in the ith interval [xi−1, xi ]. Form theRiemann sum
Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x =n∑
i=1
f (ci )∆x
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 6 / 28
Notes
Notes
Notes
2
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Theorem of the (previous) Day
Theorem
If f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{n∑
i=1
f (ci )∆x
}
exists and is the same value nomatter what choice of ci we make. x
M15 = 7.49968
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 7 / 28
Outline
Recall
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 8 / 28
The definite integral as a limit
Definition
If f is a function defined on [a, b], the definite integral of f from a to bis the number ∫ b
af (x) dx = lim
∆x→0
n∑i=1
f (ci ) ∆x
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 9 / 28
Notes
Notes
Notes
3
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Notation/Terminology
∫ b
af (x) dx = lim
∆x→0
n∑i=1
f (ci ) ∆x
I
∫— integral sign (swoopy S)
I f (x) — integrand
I a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)
I The process of computing an integral is called integration orquadrature
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 10 / 28
The limit can be simplified
Theorem
If f is continuous on [a, b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a, b]; that is, the definite integral∫ b
af (x) dx exists.
Theorem
If f is integrable on [a, b] then∫ b
af (x) dx = lim
n→∞
n∑i=1
f (xi )∆x ,
where
∆x =b − a
nand xi = a + i ∆x
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 11 / 28
Example: Integral of x
Example
Find
∫ 3
0x dx
Solution
For any n we have ∆x =3
nand xi =
3i
n. So
Rn =n∑
i=1
f (xi ) ∆x =n∑
i=1
(3i
n
)(3
n
)=
9
n2
n∑i=1
i
=9
n2· n(n + 1)
2−→ 9
2
So
∫ 3
0x dx =
9
2= 4.5
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 12 / 28
Notes
Notes
Notes
4
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Example: Integral of x2
Example
Find
∫ 3
0x2 dx
Solution
For any n we have ∆x =3
nand xi =
3i
n. So
Rn =n∑
i=1
f (xi ) ∆x =n∑
i=1
(3i
n
)2(3
n
)=
27
n3
n∑i=1
i2
=27
n3· n(n + 1)(2n + 1)
6−→ 27
3= 9
So
∫ 3
0x2 dx = 9
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 13 / 28
Example: Integral of x3
Example
Find
∫ 3
0x3 dx
Solution
For any n we have ∆x =3
nand xi =
3i
n. So
Rn =n∑
i=1
f (xi ) ∆x =n∑
i=1
(3i
n
)3(3
n
)=
81
n4
n∑i=1
i3
=81
n4· n2(n + 1)2
4−→ 81
4
So
∫ 3
0x3 dx =
81
4= 20.25
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 14 / 28
Outline
Recall
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 15 / 28
Notes
Notes
Notes
5
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Estimating the Definite Integral
Example
Estimate
∫ 1
0
4
1 + x2dx using M4.
Solution
We have x0 = 0, x1 =1
4, x2 =
1
2, x3 =
3
4, x4 = 1.
So c1 =1
8, c2 =
3
8, c3 =
5
8, c4 =
7
8.
M4 =1
4
(4
1 + (1/8)2+
4
1 + (3/8)2+
4
1 + (5/8)2+
4
1 + (7/8)2
)=
1
4
(4
65/64+
4
73/64+
4
89/64+
4
113/64
)=
64
65+
64
73+
64
89+
64
113≈ 3.1468
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 16 / 28
Estimating the Definite Integral (Continued)
Example
Estimate
∫ 1
0
4
1 + x2dx using L4 and R4
Answer
L4 =1
4
(4
1 + (0)2+
4
1 + (1/4)2+
4
1 + (1/2)2+
4
1 + (3/4)2
)= 1 +
16
17+
4
5+
16
25≈ 3.38118
R4 =1
4
(4
1 + (1/4)2+
4
1 + (1/2)2+
4
1 + (3/4)2+
4
1 + (1)2
)=
16
17+
4
5+
16
25+
1
2≈ 2.88118
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 17 / 28
Outline
Recall
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 18 / 28
Notes
Notes
Notes
6
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
1.
∫ b
ac dx = c(b − a)
2.
∫ b
a[f (x) + g(x)] dx =
∫ b
af (x) dx +
∫ b
ag(x) dx.
3.
∫ b
acf (x) dx = c
∫ b
af (x) dx.
4.
∫ b
a[f (x)− g(x)] dx =
∫ b
af (x) dx −
∫ b
ag(x) dx.
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 19 / 28
Proofs
Proofs.
I When integrating a constant function c , each Riemann sum equalsc(b − a).
I A Riemann sum for f + g equals a Riemann sum for f plus aRiemann sum for g . Using the sum rule for limits, the integral of asum is the sum of the integrals.
I Ditto for constant multiples
I Ditto for differences
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 20 / 28
Example
Find
∫ 3
0
(x3 − 4.5x2 + 5.5x + 1
)dx
Solution
∫ 3
0(x3−4.5x2 + 5.5x + 1) dx
=
∫ 3
0x3 dx − 4.5
∫ 3
0x2 dx + 5.5
∫ 3
0x dx +
∫ 3
01 dx
= 20.25− 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5
(This is the function we were estimating the integral of before)
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 21 / 28
Notes
Notes
Notes
7
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Theorem of the (previous) Day
Theorem
If f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{n∑
i=1
f (ci )∆x
}
exists and is the same value nomatter what choice of ci we make. x
M15 = 7.49968
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 22 / 28
More Properties of the Integral
Conventions: ∫ a
bf (x) dx = −
∫ b
af (x) dx
∫ a
af (x) dx = 0
This allows us to have
5.
∫ c
af (x) dx =
∫ b
af (x) dx +
∫ c
bf (x) dx for all a, b, and c.
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 23 / 28
Example
Suppose f and g are functions with
I
∫ 4
0f (x) dx = 4
I
∫ 5
0f (x) dx = 7
I
∫ 5
0g(x) dx = 3.
Find
(a)
∫ 5
0[2f (x)− g(x)] dx
(b)
∫ 5
4f (x) dx .
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 24 / 28
Notes
Notes
Notes
8
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Solution
We have
(a) ∫ 5
0[2f (x)− g(x)] dx = 2
∫ 5
0f (x) dx −
∫ 5
0g(x) dx
= 2 · 7− 3 = 11
(b) ∫ 5
4f (x) dx =
∫ 5
0f (x) dx −
∫ 4
0f (x) dx
= 7− 4 = 3
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 25 / 28
Outline
Recall
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 26 / 28
Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then∫ b
af (x) dx ≥ 0
7. If f (x) ≥ g(x) for all x in [a, b], then∫ b
af (x) dx ≥
∫ b
ag(x) dx
8. If m ≤ f (x) ≤ M for all x in [a, b], then
m(b − a) ≤∫ b
af (x) dx ≤ M(b − a)
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 27 / 28
Notes
Notes
Notes
9
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010
Example
Estimate
∫ 2
1
1
xdx using the comparison properties.
Solution
Since1
2≤ x ≤ 1
1
for all x in [1, 2], we have
1
2· 1 ≤
∫ 2
1
1
xdx ≤ 1 · 1
V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 28 / 28
Notes
Notes
Notes
10
Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010