Section 5.3 Evaluating Definite Integrals V63.0121.002.2010Su, Calculus I New York University June 21, 2010 Announcements I Final Exam is Thursday in class
Section 5.3Evaluating Definite Integrals
V63.0121.002.2010Su, Calculus I
New York University
June 21, 2010
Announcements
I Final Exam is Thursday in class
Announcements
I Sections 5.3–5.4 today
I Section 5.5 tomorrow
I Review and Movie DayWednesday
I Final exam ThursdayI roughly half-and-half
MC/FRI FR is all post-midtermI MC might have some
pre-midterm stuff on it
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 2 / 44
Resurrection Policy
If your final score beats your midterm score, we will add 10% to its weight,and subtract 10% from the midterm weight.
Image credit: Scott Beale / Laughing SquidV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 3 / 44
Objectives
I Use the Evaluation Theoremto evaluate definite integrals.
I Write antiderivatives asindefinite integrals.
I Interpret definite integrals as“net change” of a functionover an interval.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 4 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 5 / 44
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
n→∞
n∑i=1
f (ci ) ∆x
where ∆x =b − a
n, and for each i , xi = a + i∆x , and ci is a point in
[xi−1, xi ].
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 and is the same for any choice of ci .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 6 / 44
Notation/Terminology
∫ b
af (x) dx
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 7 / 44
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
Dividing up [0, 1] into 4 pieces gives
x0 = 0, x1 =1
4, x2 =
2
4, x3 =
3
4, x4 =
4
4
So the midpoint rule gives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
Dividing up [0, 1] into 4 pieces gives
x0 = 0, x1 =1
4, x2 =
2
4, x3 =
3
4, x4 =
4
4
So the midpoint rule gives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
Dividing up [0, 1] into 4 pieces gives
x0 = 0, x1 =1
4, x2 =
2
4, x3 =
3
4, x4 =
4
4
So the midpoint rule gives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
Dividing up [0, 1] into 4 pieces gives
x0 = 0, x1 =1
4, x2 =
2
4, x3 =
3
4, x4 =
4
4
So the midpoint rule gives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 9 / 44
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 10 / 44
Definite Integrals We Know So Far
I If the integral computes anarea and we know the area,we can use that. Forinstance,∫ 1
0
√1− x2 dx =
π
2
I By brute force we computed∫ 1
0x2 dx =
1
3
∫ 1
0x3 dx =
1
4
x
y
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 11 / 44
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 12 / 44
Integral of a nonnegative function is nonnegative
Proof.
If f (x) ≥ 0 for all x in [a, b], then for any number of divisions n and choiceof sample points {ci}:
Sn =n∑
i=1
f (ci )︸︷︷︸≥0
∆x ≥n∑
i=1
0 ·∆x = 0
Since Sn ≥ 0 for all n, the limit of {Sn} is nonnegative, too:∫ b
af (x) dx = lim
n→∞Sn︸︷︷︸≥0
≥ 0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 13 / 44
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 14 / 44
The definite integral is “increasing”
Proof.
Let h(x) = f (x)− g(x). If f (x) ≥ g(x) for all x in [a, b], then h(x) ≥ 0for all x in [a, b]. So by the previous property∫ b
ah(x) dx ≥ 0
This means that∫ b
af (x) dx −
∫ b
ag(x) dx =
∫ b
a(f (x)− g(x)) dx =
∫ b
ah(x) dx ≥ 0
So ∫ b
af (x) dx ≥
∫ b
ag(x) dx
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 15 / 44
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 16 / 44
Bounding the integral using bounds of the function
Proof.
If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous property∫ b
am dx ≤
∫ b
af (x) dx ≤
∫ b
aM dx
By Property 1, the integral of a constant function is the product of theconstant and the width of the interval. So:
m(b − a) ≤∫ b
af (x) dx ≤ M(b − a)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 17 / 44
Estimating an integral with inequalities
Example
Estimate
∫ 2
1
1
xdx using Property 8.
Solution
Since
1 ≤ x ≤ 2 =⇒ 1
2≤ 1
x≤ 1
1
we have1
2· (2− 1) ≤
∫ 2
1
1
xdx ≤ 1 · (2− 1)
or1
2≤∫ 2
1
1
xdx ≤ 1
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
Estimating an integral with inequalities
Example
Estimate
∫ 2
1
1
xdx using Property 8.
Solution
Since
1 ≤ x ≤ 2 =⇒ 1
2≤ 1
x≤ 1
1
we have1
2· (2− 1) ≤
∫ 2
1
1
xdx ≤ 1 · (2− 1)
or1
2≤∫ 2
1
1
xdx ≤ 1
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 19 / 44
Socratic proof
I The definite integral ofvelocity measuresdisplacement (net distance)
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 . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 20 / 44
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).
Note
In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobodyelse in the world calls it that.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
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).
Note
In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobodyelse in the world calls it that.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
Proving the Second FTC
Divide up [a, b] into n pieces of equal width ∆x =b − a
nas usual. For
each i , F is continuous on [xi−1, xi ] and differentiable on (xi−1, xi ). Sothere 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 22 / 44
Proof continued
We have for each if (ci )∆x = F (xi )− F (xi−1)
Form the Riemann Sum:
Sn =n∑
i=1
f (ci )∆x =n∑
i=1
(F (xi )− F (xi−1))
= (F (x1)− F (x0)) + (F (x2)− F (x1)) + (F (x3)− F (x2)) + · · ·· · ·+ (F (xn−1)− F (xn−2)) + (F (xn)− F (xn−1))
= F (xn)− F (x0) = F (b)− F (a)
See if you can spot the invocation of the Mean Value Theorem!
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
Proof continued
We have for each if (ci )∆x = F (xi )− F (xi−1)
Form the Riemann Sum:
Sn =n∑
i=1
f (ci )∆x =n∑
i=1
(F (xi )− F (xi−1))
= (F (x1)− F (x0)) + (F (x2)− F (x1)) + (F (x3)− F (x2)) + · · ·· · ·+ (F (xn−1)− F (xn−2)) + (F (xn)− F (xn−1))
= F (xn)− F (x0) = F (b)− F (a)
See if you can spot the invocation of the Mean Value Theorem!
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
Proof Completed
We have shown for each n,
Sn = F (b)− F (a)
so in the limit∫ b
af (x) dx = lim
n→∞Sn = lim
n→∞(F (b)− F (a)) = F (b)− F (a)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 24 / 44
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=1
4
Here we use the notation F (x)|ba or [F (x)]ba to mean F (b)− F (a).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=1
4
Here we use the notation F (x)|ba or [F (x)]ba to mean F (b)− F (a).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=1
4
Here we use the notation F (x)|ba or [F (x)]ba to mean F (b)− F (a).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and y = 1.
−1 1
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1
−1
= 2−[
1
3−(−1
3
)]=
4
3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and y = 1.
−1 1
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1
−1
= 2−[
1
3−(−1
3
)]=
4
3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and y = 1.
−1 1
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1
−1
= 2−[
1
3−(−1
3
)]=
4
3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 27 / 44
Example
Estimate
∫ 1
0
4
1 + x2dx using the midpoint rule and four divisions.
Solution
Dividing up [0, 1] into 4 pieces gives
x0 = 0, x1 =1
4, x2 =
2
4, x3 =
3
4, x4 =
4
4
So the midpoint rule gives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 28 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10
= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
Computing an integral we estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2dx .
Solution
∫ 1
0
4
1 + x2dx = 4
∫ 1
0
1
1 + x2dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π
4− 0)
= π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
Computing an integral we estimated before
Example
Evaluate
∫ 2
1
1
xdx .
Solution
∫ 2
1
1
xdx
= ln x |21
= ln 2− ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 30 / 44
Estimating an integral with inequalities
Example
Estimate
∫ 2
1
1
xdx using Property 8.
Solution
Since
1 ≤ x ≤ 2 =⇒ 1
2≤ 1
x≤ 1
1
we have1
2· (2− 1) ≤
∫ 2
1
1
xdx ≤ 1 · (2− 1)
or1
2≤∫ 2
1
1
xdx ≤ 1
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 31 / 44
Computing an integral we estimated before
Example
Evaluate
∫ 2
1
1
xdx .
Solution
∫ 2
1
1
xdx
= ln x |21
= ln 2− ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
Computing an integral we estimated before
Example
Evaluate
∫ 2
1
1
xdx .
Solution
∫ 2
1
1
xdx = ln x |21
= ln 2− ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
Computing an integral we estimated before
Example
Evaluate
∫ 2
1
1
xdx .
Solution
∫ 2
1
1
xdx = ln x |21
= ln 2− ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
Computing an integral we estimated before
Example
Evaluate
∫ 2
1
1
xdx .
Solution
∫ 2
1
1
xdx = ln x |21
= ln 2− ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 33 / 44
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 change betweenthe sides of that interval. This has many ramifications:
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
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 change betweenthe sides of that interval. This has many ramifications:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly, then∫ t1
t0
v(t) dt = s(t1)− s(t0).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
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 change betweenthe sides of that interval. This has many ramifications:
Theorem
If MC (x) represents the marginal cost of making x units of a product, then
C (x) = C (0) +
∫ x
0MC (q) dq.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
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 change betweenthe sides of that interval. This has many ramifications:
Theorem
If ρ(x) represents the density of a thin rod at a distance of x from its end,then the mass of the rod up to x is
m(x) =
∫ x
0ρ(s) ds.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 35 / 44
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation and integration,we use the indefinite integral notation∫
f (x) dx
for any function whose derivative is f (x).
Thus∫x2 dx = 1
3 x3 + C .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation and integration,we use the indefinite integral notation∫
f (x) dx
for any function whose derivative is f (x). Thus∫x2 dx = 1
3 x3 + C .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 37 / 44
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 38 / 44
Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and thevertical lines x = 0 and x = 3.
Solution
Consider
∫ 3
0(x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
and (2, 3], and negative on (1, 2). If we want the area of the region, wehave to do
A =
∫ 1
0(x − 1)(x − 2) dx −
∫ 2
1(x − 1)(x − 2) dx +
∫ 3
2(x − 1)(x − 2) dx
=[
13 x3 − 3
2 x2 + 2x]1
0−[
13 x3 − 3
2 x2 + 2x]2
1+[
13 x3 − 3
2 x2 + 2x]3
2
=5
6−(−1
6
)+
5
6=
11
6.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 39 / 44
Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and thevertical lines x = 0 and x = 3.
Solution
Consider
∫ 3
0(x − 1)(x − 2) dx.
Notice the integrand is positive on [0, 1)
and (2, 3], and negative on (1, 2). If we want the area of the region, wehave to do
A =
∫ 1
0(x − 1)(x − 2) dx −
∫ 2
1(x − 1)(x − 2) dx +
∫ 3
2(x − 1)(x − 2) dx
=[
13 x3 − 3
2 x2 + 2x]1
0−[
13 x3 − 3
2 x2 + 2x]2
1+[
13 x3 − 3
2 x2 + 2x]3
2
=5
6−(−1
6
)+
5
6=
11
6.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 39 / 44
Graph
x
y
1 2 3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 40 / 44
Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and thevertical lines x = 0 and x = 3.
Solution
Consider
∫ 3
0(x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
and (2, 3], and negative on (1, 2).
If we want the area of the region, wehave to do
A =
∫ 1
0(x − 1)(x − 2) dx −
∫ 2
1(x − 1)(x − 2) dx +
∫ 3
2(x − 1)(x − 2) dx
=[
13 x3 − 3
2 x2 + 2x]1
0−[
13 x3 − 3
2 x2 + 2x]2
1+[
13 x3 − 3
2 x2 + 2x]3
2
=5
6−(−1
6
)+
5
6=
11
6.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and thevertical lines x = 0 and x = 3.
Solution
Consider
∫ 3
0(x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
and (2, 3], and negative on (1, 2). If we want the area of the region, wehave to do
A =
∫ 1
0(x − 1)(x − 2) dx −
∫ 2
1(x − 1)(x − 2) dx +
∫ 3
2(x − 1)(x − 2) dx
=[
13 x3 − 3
2 x2 + 2x]1
0−[
13 x3 − 3
2 x2 + 2x]2
1+[
13 x3 − 3
2 x2 + 2x]3
2
=5
6−(−1
6
)+
5
6=
11
6.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
Interpretation of “negative area” in motion
There is an analog in rectlinear motion:
I
∫ t1
t0
v(t) dt is net distance traveled.
I
∫ t1
t0
|v(t)| dt is total distance traveled.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 42 / 44
What about the constant?
I It seems we forgot about the +C when we say for instance∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=1
4− 0 =
1
4
I But notice[x4
4+ C
]1
0
=
(1
4+ C
)− (0 + C ) =
1
4+ C − C =
1
4
no matter what C is.
I So in antidifferentiation for definite integrals, the constant isimmaterial.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 43 / 44
Summary
I The second Fundamental Theorem of Calculus:∫ b
af (x) dx = F (b)− F (a)
where F ′ = f .
I Definite integrals represent net change of a function over an interval.
I We write antiderivatives as indefinite integrals
∫f (x) dx
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 44 / 44