Section 5.3
Evaluating Definite Integrals
V63.0121, Calculus I
April 20, 2009
Announcements
Final Exam is Friday, May 8, 2:00–3:50pm
Final is cumulative; topics will be represented roughly
according to time spent on them
.
.
Image credit: docman
. . . . . .

Outline
Last time: The Definite Integral
Evaluating Definite Integrals
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”
. . . . . .

The definite integral as a limit
Definition
If f is a function 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
n→∞
a i=1
b−a
, and for each i, xi = a + i∆x, and ci is a point
where ∆x =
n
in [xi−1 , xi ].
. . . . . .

Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b
the upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration
. . . . . .

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
∫b
c dx = c(b − a)
1.
a
∫ ∫ ∫
b b b
[f(x) + g(x)] dx = f(x) dx + g(x) dx.
2.
a a a
∫ ∫
b b
cf(x) dx = c f(x) dx.
3.
a a
∫ ∫ ∫
b b b
[f(x) − g(x)] dx = f(x) dx − g(x) dx.
4.
a a a
. . . . . .

More Properties of the Integral
Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
∫c ∫b ∫ c
f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
5.
a a b
. . . . . .

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
f(x) dx ≥ 0
a
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ ∫
b b
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
. . . . . .

Outline
Last time: The Definite Integral
Evaluating Definite Integrals
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”
. . . . . .

Socratic proof
The definite integral of
velocity measures
displacement (net
distance)
The derivative of
displacement is velocity
So we can compute
displacement with the
antiderivative of
velocity?
. . . . . .

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
f(x) dx = F(b) − F(a).
a
. . . . . .

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
f(x) dx = F(b) − F(a).
a
Note
In Section 5.3., this theorem is called “The Evaluation Theorem”.
Nobody else in the world calls it that.
. . . . . .

Proving 2FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as
n
usual. 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 )
= F′ (ci ) = f(ci )
xi − xi−1
Or
f(ci )∆x = F(xi ) − F(xi−1 )
. . . . . .

We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
n n
∑ ∑
(F(xi ) − F(xi−1 ))
Sn = f(ci )∆x =
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−1 )) + (F(xn ) − F(xn−1 ))
= F(xn ) − F(x0 ) = F(b) − F(a)
. . . . . .

We have shown for each n,
Sn = F(b) − F(a)
so in the limit
∫b
f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a)
n→∞ n→∞
a
. . . . . .

Example
Find the area between y = x3 and the x-axis, between x = 0 and
x = 1.
.
. . . . . .

Example
Find the area between y = x3 and the x-axis, between x = 0 and
x = 1.
Solution
∫ 1
1
x4 1
x3 dx =
A= =
4 4 .
0 0
. . . . . .

Example
Find the area between y = x3 and the x-axis, between x = 0 and
x = 1.
Solution
∫ 1
1
x4 1
x3 dx =
A= =
4 4 .
0 0
Here we use the notation F(x)|b or [F(x)]b to mean F(b) − F(a).
a a
. . . . . .

Example
Find the area enclosed by the parabola y = x2 and y = 1.
. . . . . .

Example
Find the area enclosed by the parabola y = x2 and y = 1.
.
. . . . . .

Example
Find the area enclosed by the parabola y = x2 and y = 1.
.
Solution
[ ]1 [ ( )]
∫ 1
x3 −1
1 4
2
A=2− x dx = 2 − =2− −− =
3 3 3 3
−1 −1
. . . . . .

Outline
Last time: The Definite Integral
Evaluating Definite Integrals
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”
. . . . . .

The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
. . . . . .

The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly,
then ∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
. . . . . .

The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
Theorem
If MC(x) represents the marginal cost of making x units of a
product, then
∫x
C(x) = C(0) + MC(q) dq.
0
. . . . . .

The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the 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
∫x
m(x) = ρ(s) ds.
0
. . . . . .

Outline
Last time: The Definite Integral
Evaluating Definite Integrals
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”
. . . . . .

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).
. . . . . .

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 x3 + C.
3
. . . . . .

My first table of integrals
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫
∫
xn+1
xn dx = cf(x) dx = c f(x) dx
+ C (n ̸= −1)
n+1 ∫
∫
1
ex dx = ex + C dx = ln |x| + C
x
∫
∫
ax
ax dx = +C
sin x dx = − cos x + C
ln a
∫
∫
csc2 x dx = − cot x + C
cos x dx = sin x + C
∫
∫
sec2 x dx = tan x + C csc x cot x dx = − csc x + C
∫
∫
1
√ dx = arcsin x + C
sec x tan x dx = sec x + C
1 − x2
∫
1
dx = arctan x + C
1 + x2
. . . . . .

Outline
Last time: The Definite Integral
Evaluating Definite Integrals
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”
. . . . . .

Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis,
and the vertical lines x = 0 and x = 3.
. . . . . .

Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis,
and the vertical lines x = 0 and x = 3.
Solution ∫
3
(x − 1)(x − 2) dx. Notice the integrand is positive on
Consider
0
[0, 1) and (2, 3], and negative on (1, 2). If we want the area of
the region, we have to do
∫ ∫ ∫
1 2 3
(x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
A=
0 1 2
[1 ]1 [1 3 ]2 [1 ]3
x3 − 3 x2 + 2x 0 − − 3 x2 + 2x 3 32
3 x − 2 x + 2x
3x
= +
3
(2) 2 1 2
5 1 5 11
= −− += .
6 6 6 6
. . . . . .

Graph from previous example
y
.
. . . . x
.
2
. 3
.
1
.
. . . . . .

Summary
integrals can be computed with antidifferentiation
integral of instantaneous rate of change is total net change
The second Funamental Theorem of Calculus requires the
Mean Value Theorem
. . . . . .