final exam review (integration)
DESCRIPTION
review for the integration portion of the final exam.TRANSCRIPT
Review for Final ExamIntegration
Math 1a
January 13, 2008
Announcements
I Office hours on the website (click “Exams”)
I Email your TF, CA, or me with questions
I Final: Thursday 9:15am in Hall B
Outline
The Riemann IntegralEstimating the integralProperties of the integralComparison Properties of the Integral
The Fundamental Theorem of CalculusStatementDifferentiation of functions defined by integralsProperties of the area functionThe Second FTCExamplesTotal ChangeIndefinite Integrals
Integration by SubstitutionSubstitution for Indefinite IntegralsSubstitution for Definite Integrals
The Riemann IntegralLearning Objectives
I Compute the definite integral using a limit of Riemann sums
I Estimate the definite integral using a Riemann sum (e.g.,Midpoint Rule)
I Reason with the definite integral using its elementaryproperties.
The Area Problem
Given a function f defined on [a, b], how can one find the areabetween y = 0, y = f (x), x = a, and x = b?We divide and conquer.
Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f (x0)∆x + f (x1)∆x + · · ·+ f (xn−1)∆x
Rn = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x
Mn = f
(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · ·+ f
(xn−1 + xn
2
)∆x
In general, choose ci to be a point in the ith interval [xi−1, xi ].Form the Riemann sum
Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x
=n∑
i=1
f (ci )∆x
Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f (x0)∆x + f (x1)∆x + · · ·+ f (xn−1)∆x
Rn = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x
Mn = f
(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · ·+ f
(xn−1 + xn
2
)∆x
In general, choose ci to be a point in the ith interval [xi−1, xi ].Form the Riemann sum
Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x
=n∑
i=1
f (ci )∆x
Theorem
TheoremIf f is a continuous function on [a, b] or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x}
exists and is the same value no matter what choice of ci we made.
DefinitionThe definite integral of f from a to b is the number∫ b
af (x) dx = lim
∆x→0
n∑i=1
f (ci ) ∆x
Theorem
TheoremIf f is a continuous function on [a, b] or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x}
exists and is the same value no matter what choice of ci we made.
DefinitionThe definite integral of f from a to b is the number∫ b
af (x) dx = lim
∆x→0
n∑i=1
f (ci ) ∆x
Example (“Sample Exam”, Problem 6)
The rate at which the world’s oil is being consumed is increasing.Suppose that the rate (measured in billions of barrels per year) isgiven by the function r(t), where t is measured in years and t = 0represents January 1, 2000.
(a) Write a definite integral that represents the total quantity ofoil used between the start of 2000 and the start of 2005.
(b) Suppose that r(t) = 32e0.05t . Find the approximate value forthe definite integral from part (a) using a right-hand sum withn = 5 subintervals.
(c) Interpret each of the five terms in the sum from part (b) interms of oil consumption.
Answers
(a)
∫ 5
0r(t) dt
(b)
1·32e0.05(1)+1·32e0.05(2)+1·32e0.05(3)+1·32e0.05(4)+1·32e0.05(5)
(c) Each term stands for the approximate amount of oil used ineach year. For instance, the term 1 · 32e0.05(3) is approximatelythe amount of oil used between January 1, 2002 and January1, 2003.
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
The partition is 0 <1
4<
1
2<
3
4< 1, so the estimate is
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
)=
150, 166, 784
47, 720, 465≈ 3.1468
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
The partition is 0 <1
4<
1
2<
3
4< 1, so the estimate is
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
)=
150, 166, 784
47, 720, 465≈ 3.1468
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
The partition is 0 <1
4<
1
2<
3
4< 1, so the estimate is
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
)
=150, 166, 784
47, 720, 465≈ 3.1468
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
The partition is 0 <1
4<
1
2<
3
4< 1, so the estimate is
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
)=
150, 166, 784
47, 720, 465≈ 3.1468
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.
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.
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.
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.
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 .
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 .
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 .
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 .
SolutionWe 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
SolutionWe 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
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)
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)
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)
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)
Example
Estimate
∫ 4
1
1√x + sin2 πx
dx using the comparison properties.
Outline
The Riemann IntegralEstimating the integralProperties of the integralComparison Properties of the Integral
The Fundamental Theorem of CalculusStatementDifferentiation of functions defined by integralsProperties of the area functionThe Second FTCExamplesTotal ChangeIndefinite Integrals
Integration by SubstitutionSubstitution for Indefinite IntegralsSubstitution for Definite Integrals
The Fundamental Theorem of CalculusLearning Objectives
I State and use both fundamental theorems of calculus
I Understand the relationship between integration andantidifferentiation
I Use FTC to compute derivatives of integrals with functions inthe limits
I Use FTC to compute areas or other accumulations
Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable function on [a, b] and define
g(x) =
∫ x
af (t) dt.
If f is continuous at x in (a, b), then g is differentiable at x and
g ′(x) = f (x).
Example (Spring 2000 Final, Problem 7c)
Finddy
dxif y =
∫ 100
x3+x
√p2 − p dp
Solution
Let A(u) =
∫ u
1
√p2 − p dp. By the Fundamental Theorem of
Calculus, A′(u) =√
u2 − u. We have
y ′ =d
dx
[∫ 100
x3+x
√p2 − p dp
]=
d
dx
[∫ 100
1
√p2 − p dp −
∫ x3+x
1
√p2 − p dp
]
=d
dx
[A(100)− A(x3 + x)
]= −A′(x3 + x) · (3x2 + 1)
= −(3x2 + 1)√
(x3 + x)2 − (x3 + x).
Example (Spring 2000 Final, Problem 7c)
Finddy
dxif y =
∫ 100
x3+x
√p2 − p dp
Solution
Let A(u) =
∫ u
1
√p2 − p dp. By the Fundamental Theorem of
Calculus, A′(u) =√
u2 − u. We have
y ′ =d
dx
[∫ 100
x3+x
√p2 − p dp
]=
d
dx
[∫ 100
1
√p2 − p dp −
∫ x3+x
1
√p2 − p dp
]
=d
dx
[A(100)− A(x3 + x)
]= −A′(x3 + x) · (3x2 + 1)
= −(3x2 + 1)√
(x3 + x)2 − (x3 + x).
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s velocityat time t = 5?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s velocityat time t = 5?
SolutionRecall that by the FTC wehave
s ′(t) = f (t).
So s ′(5) = f (5) = 2.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Is the acceleration of the par-ticle at time t = 5 positive ornegative?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Is the acceleration of the par-ticle at time t = 5 positive ornegative?
SolutionWe have s ′′(5) = f ′(5), whichlooks negative from thegraph.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s positionat time t = 3?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
What is the particle’s positionat time t = 3?
SolutionSince on [0, 3], f (x) = x, wehave
s(3) =
∫ 3
0x dx =
9
2.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
Solution
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionThe critical points of s arethe zeros of s ′ = f .
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionBy looking at the graph, wesee that f is positive fromt = 0 to t = 6, then negativefrom t = 6 to t = 9.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
At what time during the first 9seconds does s have its largestvalue?
SolutionTherefore s is increasing on[0, 6], then decreasing on[6, 9]. So its largest value isat t = 6.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Approximately when is the ac-celeration zero?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
Approximately when is the ac-celeration zero?
Solutions ′′ = 0 when f ′ = 0, whichhappens at t = 4 and t = 7.5(approximately)
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionThe particle is moving awayfrom the origin when s > 0and s ′ > 0.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionSince s(0) = 0 and s ′ > 0 on(0, 6), we know the particle ismoving away from the originthen.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
When is the particle movingtoward the origin? Away fromthe origin?
SolutionAfter t = 6, s ′ < 0, so theparticle is moving toward theorigin.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
On which side (positive or neg-ative) of the origin does theparticle lie at time t = 9?
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
On which side (positive or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionWe have s(9) =∫ 6
0f (x) dx +
∫ 9
6f (x) dx,
where the left integral ispositive and the right integralis negative.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
On which side (positive or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionIn order to decide whethers(9) is positive or negative,we need to decide if the firstarea is more positive than thesecond area is negative.
Facts about g from f
Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving
along a coordinate axis is s(t) =
∫ t
0f (x) dx meters. Use the
graph to answer the following questions.
1 2 3 4 5 6 7 8 9
1
2
3
4
• (1,1)
• (2,2)
• (3,3)• (5,2)
On which side (positive or neg-ative) of the origin does theparticle lie at time t = 9?
SolutionThis appears to be the case,so s(9) is positive.
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).
Examples
Find the following integrals:
I
∫ 1
0x2 dx ,
∫ 1
0x3 dx ,
∫ 2
1xn dx (n 6= −1),
∫ 2
1
1
xdx
I
∫ π
0sin θ dθ,
∫ 1
0ex dx
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf v(t) represents the velocity of a particle moving rectilinearly,then ∫ t1
t0
v(t) dt = s(t1)− s(t0).
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf MC (x) represents the marginal cost of making x units of aproduct, then
C (x) = C (0) +
∫ x
0MC (q) dq.
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf ρ(x) represents the density of a thin rod at a distance of x fromits end, then the mass of the rod up to x is
m(x) =
∫ x
0ρ(s) ds.
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫
f (x) dx
for any function whose derivative is f (x).
Thus∫x2 dx = 1
3 x3 + C .
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫
f (x) dx
for any function whose derivative is f (x). Thus∫x2 dx = 1
3 x3 + C .
My first table of integrals∫[f (x) + g(x)] dx =
∫f (x) dx +
∫g(x) dx∫
xn dx =xn+1
n + 1+ C (n 6= −1)∫
ex dx = ex + C∫sin x dx = − cos x + C∫cos x dx = sin x + C∫
sec2 x dx = tan x + C∫sec x tan x dx = sec x + C∫
1
1 + x2dx = arctan x + C
∫cf (x) dx = c
∫f (x) dx∫
1
xdx = ln |x |+ C∫
ax dx =ax
ln a+ C∫
csc2 x dx = − cot x + C∫csc x cot x dx = − csc x + C∫
1√1− x2
dx = arcsin x + C
Outline
The Riemann IntegralEstimating the integralProperties of the integralComparison Properties of the Integral
The Fundamental Theorem of CalculusStatementDifferentiation of functions defined by integralsProperties of the area functionThe Second FTCExamplesTotal ChangeIndefinite Integrals
Integration by SubstitutionSubstitution for Indefinite IntegralsSubstitution for Definite Integrals
Integration by SubstitutionLearning Objectives
I Given an integral and a specific substitution, perform thatsubstitution
I Use the substitution method to evaluate definite andindefinite integrals
Theorem (The Substitution Rule)
If u = g(x) is a differentiable function whose range is an interval Iand f is continuous on I , then∫
f (g(x))g ′(x) dx =
∫f (u) du
or ∫f (u)
du
dxdx =
∫f (u) du
This is the “anti” version of the chain rule.
Example
Find
∫xex2
dx
SolutionLet u = x2. Then du = 2x dx and x dx = 1
2 du. So∫xex2
dx = 12
∫eu du
= 12 eu + C
= 12 ex2
+ C
Example
Find
∫xex2
dx
SolutionLet u = x2. Then du = 2x dx and x dx = 1
2 du. So∫xex2
dx = 12
∫eu du
= 12 eu + C
= 12 ex2
+ C
Theorem (The Substitution Rule for Definite Integrals)
If g ′ is continuous and f is continuous on the range of u = g(x),then ∫ b
af (g(x))g ′(x) dx =
∫ g(b)
g(a)f (u) du.
Example
Compute
∫ π
0cos2 x sin x dx .
Solution (Slow Way)
First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −
∫u2 du
= −13 u3 + C = −1
3 cos3 x + C .
Therefore ∫ π
0cos2 x sin x dx = −1
3 cos3 x∣∣π0
= 23 .
Example
Compute
∫ π
0cos2 x sin x dx .
Solution (Slow Way)
First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.
Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −
∫u2 du
= −13 u3 + C = −1
3 cos3 x + C .
Therefore ∫ π
0cos2 x sin x dx = −1
3 cos3 x∣∣π0
= 23 .
Example
Compute
∫ π
0cos2 x sin x dx .
Solution (Slow Way)
First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −
∫u2 du
= −13 u3 + C = −1
3 cos3 x + C .
Therefore ∫ π
0cos2 x sin x dx = −1
3 cos3 x∣∣π0
= 23 .
Solution (Fast Way)
Do both the substitution and the evaluation at the same time.
Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π
0cos2 x sin x dx =
∫ −1
1−u2 du
=
∫ 1
−1u2 du
= 13 u3∣∣1−1
=2
3.
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.
So∫ π
0cos2 x sin x dx =
∫ −1
1−u2 du
=
∫ 1
−1u2 du
= 13 u3∣∣1−1
=2
3.
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π
0cos2 x sin x dx =
∫ −1
1−u2 du
=
∫ 1
−1u2 du
= 13 u3∣∣1−1
=2
3.