Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental Theorem of Calculus!

1. 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
. . . . . .

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

3. 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 ].
. . . . . .

4. 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
. . . . . .

5. 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
. . . . . .

6. 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
. . . . . .

7. 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
. . . . . .

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

9. 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?
. . . . . .

10. 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
. . . . . .

11. 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.
. . . . . .

12. 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 )
. . . . . .

13. 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)
. . . . . .

14. 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
. . . . . .

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

16. 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
. . . . . .

17. 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
. . . . . .

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

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

20. 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
. . . . . .

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

22. 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:
. . . . . .

23. 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
. . . . . .

24. 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
. . . . . .

25. 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
. . . . . .

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

27. 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).
. . . . . .

28. 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
. . . . . .

29. 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
. . . . . .

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

31. 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.
. . . . . .

32. 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
. . . . . .

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

34. 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
. . . . . .