The Second Fundamental Theorem of Calculus allows us to compute definite integrals by antidifferentiation

1. Section 5.3
Evaluating Definite Integrals
Math 1a
December 10, 2007
Announcements
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2. Outline
FTC2
Proof
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”

3. Theorem (The Second Fundamental Theorem of Calculus,
Strong Form)
Suppose f is integrable on [a, b] and f = F for another function f ,
then
b
f (x) dx = F (b) − F (a).
a

4. Proof.
We will choose Riemann sums which converge to the right-hand
side. Let n be given. On the interval [xi−1 , xi ] there is a point ci
such that
F (xi ) − F (xi−1 )
f (ci ) = F (ci ) =
xi − xi−1
Then
n n
F (xi ) − F (xi−1 )
Rn = f (ci )∆x =
i=1 i=1
= F (xn ) − F (x0 ) = F (b) − F (a).
So Rn → F (b) − F (a) as n → ∞.

5. Examples
Find the following integrals:
1 1 1 2
1
x 2 dx, x 3 dx, x n dx (n = −1), dx
x
0 0 0 1
π 1 π/4
e x dx,
sin θ dθ, tan θ dθ
0 0 0

6. Outline
FTC2
Proof
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”

7. 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:

8. 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

9. 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

10. 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

11. Example
If oil leaks from a tank at a rate of r (t) gallons per minute at time
120
t, what does r (t) dt represent?
0

12. Example
If oil leaks from a tank at a rate of r (t) gallons per minute at time
120
t, what does r (t) dt represent?
0
Solution
The amount of oil lost in two hours.

13. Outline
FTC2
Proof
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”

14. 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).

15. 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
x 2 dx = 3 x 3 + C .
1

16. My first table of integrals
[f (x) + g (x)] dx = f (x) dx + g (x) dx
x n+1
x n dx = cf (x) dx = c f (x) dx
+ C (n = −1)
n+1
1
e x dx = e x + 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

17. Outline
FTC2
Proof
Examples
Total Change
Indefinite Integrals
My first table of integrals
Examples
“Negative Area”

18. 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.

19. 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 2 3
13
− 2 x 2 + 2x 13
− 2 x 2 + 2x 13
− 2 x 2 + 2x
3 3 3
−
= 3x 3x + 3x
0 1 2
5 1 5 11
−−
= + =.
6 6 6 6