. . Section 5.3 Evaluating Definite Integrals V63.0121.041, Calculus I New York University December 6, 2010 Announcements I Today: Section 5.3 I Wednesday: Section 5.4 I Monday, December 13: Section 5.5 I ”Monday,” December 15: Review and Movie Day! I Monday, December 20, 12:00–1:50pm: Final Exam . . . . . .
A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.
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Section 5.3Evaluating Definite Integrals
V63.0121.041, Calculus I
New York University
December 6, 2010
Announcements
I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13: Section 5.5I ”Monday,” December 15: Review and Movie Day!I Monday, December 20, 12:00–1:50pm: Final Exam
. . . . . .
. . . . . .
Announcements
I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13:Section 5.5
I ”Monday,” December 15:Review and Movie Day!
I Monday, December 20,12:00–1:50pm: Final Exam
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41
. . . . . .
Objectives
I Use the EvaluationTheorem to evaluatedefinite integrals.
I Write antiderivatives asindefinite integrals.
I Interpret definite integralsas “net change” of afunction over an interval.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 3 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 4 / 41
. . . . . .
The definite integral as a limit
DefinitionIf f is a function defined on [a,b], the definite integral of f from a to bis the number ∫ b
af(x)dx = lim
n→∞
n∑i=1
f(ci)∆x
where ∆x =b− an
, and for each i, xi = a+ i∆x, and ci is a point in[xi−1, xi].
TheoremIf 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 and is the same for any choice of ci.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 5 / 41
. . . . . .
Notation/Terminology
∫ b
af(x)dx
I∫
— integral sign (swoopy S)
I f(x) — integrandI 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
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 6 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=
14
(4
65/64+
473/64
+4
89/64+
4113/64
)
=150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=
14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 8 / 41
. . . . . .
More Properties of the Integral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx∫ a
af(x)dx = 0
This allows us to have
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 9 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
.
∫ c
bf(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
.
∫ c
bf(x)dx
.
∫ c
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ b
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ c
bf(x)dx =
−∫ b
cf(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ c
bf(x)dx =
−∫ b
cf(x)dx
.
∫ c
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Definite Integrals We Know So Far
I If the integral computes anarea and we know thearea, we can use that. Forinstance,∫ 1
0
√1− x2 dx =
π
4
I By brute force wecomputed∫ 1
0x2 dx =
13
∫ 1
0x3 dx =
14
..x
.
y
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 11 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet 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)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet 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)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet 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)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Estimating an integral with inequalities
Example
Estimate∫ 2
1
1xdx using Property 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1we have
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
. . . . . .
Estimating an integral with inequalities
Example
Estimate∫ 2
1
1xdx using Property 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1we have
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 14 / 41
. . . . . .
Socratic proof
I The definite integral ofvelocity measuresdisplacement (netdistance)
I The derivative ofdisplacement is velocity
I So we can computedisplacement with thedefinite integral or theantiderivative of velocity
I But any function can be avelocity function, so . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 15 / 41
. . . . . .
Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a,b] and f = F′ for another function F, then∫ b
af(x)dx = F(b)− F(a).
NoteIn Section 5.3, this theorem is called “The Evaluation Theorem”.Nobody else in the world calls it that.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
. . . . . .
Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a,b] and f = F′ for another function F, then∫ b
af(x)dx = F(b)− F(a).
NoteIn Section 5.3, this theorem is called “The Evaluation Theorem”.Nobody else in the world calls it that.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41