A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

1. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Sec on 5.3
Evalua ng Definite Integrals
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 27, 2011
.
.
Notes
Announcements
Today: 5.3
Thursday/Friday: Quiz on
4.1–4.4
Monday 5/2: 5.4
Wednesday 5/4: 5.5
Monday 5/9: Review and
Movie Day!
Thursday 5/12: Final
Exam, 2:00–3:50pm
.
.
Notes
Objectives
Use the Evalua on
Theorem to evaluate
definite integrals.
Write an deriva ves as
indefinite integrals.
Interpret definite
integrals as “net change”
of a func on over an
interval.
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. 1
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2. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals
.
.
Notes
The definite integral as a limit
Defini on
If f is a func on defined on [a, b], the definite integral of f from a to
b is the number
∫ b ∑n
f(x) dx = lim f(ci ) ∆x
a n→∞
i=1
b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].
.
.
Notes
The definite integral as a limit
Theorem
If f is con nuous on [a, b] or if f has only finitely many jump
discon nui es, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists and is the same for any choice of ci .
a
.
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. 2
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3. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integra on (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of compu ng an integral is called integra on
.
.
Example
∫
Notes
1
4
Es mate dx using M4 .
0 1 + x2
Solu on
1 1 3
We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1.
4 2 4
1 3 5 7
So c1 = , c2 = , c3 = , c4 = .
8 8 8 8
.
.
Example
∫
Notes
1
4
Es mate dx using M4 .
0 1 + x2
Solu on
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
( )
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
64 64 64 64
= + + + ≈ 3.1468
65 73 89 113
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. 3
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4. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Properties of the integral
Theorem (Addi ve Proper es of the Integral)
Let f and g be integrable func ons on [a, b] and c a constant. Then
∫ b
1. c dx = c(b − a)
∫a b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
∫a b ∫ b a a
3. cf(x) dx = c f(x) dx.
∫a b a
∫ b ∫ b
4. [f(x) − g(x)] dx = f(x) dx − g(x) dx.
a a a
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Notes
More Properties of the Integral
Conven ons: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
.
.
Notes
Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b ∫ c
f(x) dx f(x) dx
a b
.
a c x
b
.
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. 4
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5. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ ∫ c
c
f(x) dx f(x) dx =
b∫
a b
− f(x) dx
. c
a c x
b
.
.
Notes
Definite Integrals We Know So Far
If the integral computes an area
and we know the area, we can
use that. For instance,
∫ 1√ y
π
1 − x2 dx =
0 4
By brute force we computed .
∫ 1 ∫ 1 x
1 1
x2 dx = x3 dx =
0 3 0 4
.
.
Notes
Comparison Properties of the Integral
Theorem
Let f and g be integrable func ons on [a, b].
∫ b
6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0
a
∫ b ∫ b
7. If f(x) ≥ g(x) for all x in [a, b], then f(x) dx ≥ g(x) dx
a a
8. If m ≤ f(x) ≤ M for all x in [a, b], then
∫ b
m(b − a) ≤ f(x) dx ≤ M(b − a)
a
.
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. 5
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6. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
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 choice
of sample points {ci }:
∑
n ∑
n
Sn = f(ci ) ∆x ≥ 0 · ∆x = 0
i=1 ≥0 i=1
. x
Since Sn ≥ 0 for all n, the limit of {Sn } is nonnega ve, too:
∫ b
f(x) dx = lim Sn ≥ 0
a n→∞
≥0
.
.
Notes
The 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) ≥ 0 for all f(x)
x in [a, b]. So by the previous h(x) g(x)
property
∫ b
h(x) dx ≥ 0 . x
a
This means that
∫ b ∫ b ∫ b ∫ b
f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0
a a a a
.
.
Notes
Bounding the integral
Proof.
If m ≤ f(x) ≤ M on for all x in [a, b], then by
y
the previous property
∫ b ∫ b ∫ b M
m dx ≤ f(x) dx ≤ M dx
a a a f(x)
By Property 8, the integral of a constant
func on is the product of the constant and m
the width of the interval. So:
∫ b . x
m(b − a) ≤ f(x) dx ≤ M(b − a) a b
a
.
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. 6
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7. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Example Notes
∫ 2
1
Es mate dx using the comparison proper es.
1 x
Solu on
.
.
Notes
Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals
.
.
Notes
Socratic proof
The definite integral of velocity
measures displacement (net
distance)
The deriva ve of displacement
is velocity
So we can compute
displacement with the definite
integral or the an deriva ve of
velocity
But any func on can be a
velocity func on, so . . .
.
.
. 7
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8. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another func on F,
then ∫ b
f(x) dx = F(b) − F(a).
a
.
.
Notes
Proving the Second FTC
Proof.
b−a
Divide up [a, b] into n pieces of equal width ∆x = as
n
usual.
For each i, F is con nuous on [xi−1 , xi ] and differen able on
(xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with
F(xi ) − F(xi−1 )
= F′ (ci ) = f(ci )
xi − xi−1
=⇒ f(ci )∆x = F(xi ) − F(xi−1 )
.
.
Notes
Proving the Second FTC
Proof.
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=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)
.
.
. 8
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9. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Proving the Second FTC
Proof.
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.
So in the limit
∫ b
f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a)
a n→∞ n→∞
.
.
Notes
Computing area with the 2nd FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and
x = 1.
Solu on
∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .
Here we use the nota on F(x)|b or [F(x)]b to mean F(b) − F(a).
a a
.
.
Notes
Computing area with the 2nd FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
Solu on
∫ 1 [ 3 ]1
x
A=2− x2 dx = 2 − 1
−1 3 −1
[ ( )]
1 1 4
=2− − − = .
3 3 3
−1 1
.
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. 9
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10. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Computing an integral we Notes
estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solu on
.
.
Computing an integral we Notes
estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solu on
.
.
Notes
Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals
.
.
. 10
.

11. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
The Integral as Net Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a deriva ve along an interval is the net change
over that interval. This has many interpreta ons.
.
.
Notes
The Integral as Net Change
Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
.
.
Notes
The Integral as Net Change
Corollary
If MC(x) represents the marginal cost of making x units of a product,
then ∫ x
C(x) = C(0) + MC(q) dq.
0
.
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. 11
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12. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
The Integral as Net Change
Corollary
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
∫ x
m(x) = ρ(s) ds.
0
.
.
Notes
Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals
.
.
Notes
A new notation for antiderivatives
To emphasize the rela onship between an differen a on and
integra on, we use the indefinite integral nota on
∫
f(x) dx
for any func on whose deriva ve is f(x). Thus
∫
x2 dx = 1 x3 + C.
3
.
.
. 12
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13. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
My first table of integrals
.
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+1
xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
∫ n+1 ∫
1
ex dx = ex + C dx = ln |x| + C
∫ ∫ x
x ax
sin x dx = − cos x + C a dx = +C
∫ ln a
∫
cos x dx = sin x + C csc2 x dx = − cot x + C
∫ ∫
sec2 x dx = tan x + C csc x cot x dx = − csc x + C
∫ ∫
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
∫ 1 − x2
1
dx = arctan x + C
1 + x2
.
.
Notes
Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals
.
.
Notes
Computing Area with integrals
,label=area-exp
Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .
,label=area-exp
Solu on
The answer is ∫ 4
ex dx = ex |4 = e4 − e.
1
1
.
.
. 13
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14. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solu on
.
.
Example
Notes
Find the area between the graph of y = (x − 1)(x − 2), the x-axis,
and the ver cal lines x = 0 and x = 3.
Solu on
.
.
Interpretation of “negative area” Notes
in motion
There is an analog in rectlinear mo on:
∫ t1
v(t) dt is net distance traveled.
t
∫ 0t1
|v(t)| dt is total distance traveled.
t0
.
.
. 14
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15. . V63.0121.001: Calculus I
. Sec on 5.3: Evalua ng Definite Integrals
. April 27, 2011
Notes
What about the constant?
It seems we forgot about the +C when we say for instance
∫ 1 1
x4 1 1
x3 dx = = −0=
0 4 0 4 4
But no ce
[ 4 ]1 ( )
x 1 1 1
+C = + C − (0 + C) = + C − C =
4 0 4 4 4
no ma er what C is.
So in an differen a on for definite integrals, the constant is
immaterial.
.
.
Notes
Summary
The second Fundamental Theorem of Calculus:
∫ b
f(x) dx = F(b) − F(a)
a
′
where F = f.
Definite integrals represent net change of a func on over an
interval. ∫
We write an deriva ves as indefinite integrals f(x) dx
.
.
Notes
.
.
. 15
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