# Document 10434907 ```MATH 131-503 Fall 2015
c
Wen
Liu
5.3
5.3 Evaluating Definite Integrals
Evaluation Theorem: If f is continuous on the interval [a, b], then
Z
b
f (x)dx = F (b) − F (a)
a
where F is any antiderivative of f , that is, F 0 = f .
Table of Indefinite Integrals
Z
•
Z
cf (x)dx = c
Z
•
Z
•
Z
f (x)dx
xn+1
x dx =
+ C, n 6= −1
n+1
n
x
x
e dx = e + C
sin xdx = − cos x + C
Z
•
2
sec xdx = tan x + C
sec x tan xdx = sec x + C
Z
•
Z
•
Z
•
f (x)dx &plusmn;
g(x)dx
1
dx = ln |x| + C
x
ax dx =
ax
+C
ln a
•
cos xdx = sin x + C
Z
•
csc2 xdx = − cot x + C
Z
Z
•
(f (x) &plusmn; g(x)) dx =
Z
Z
Z
•
•
Z
1
dx = tan−1 x + C
2
x +1
•
csc x cot xdx = − csc x + C
Z
•
1
√
dx = sin−1 x + C
2
1−x
Examples: Compute each of the following.
Z
1
3
1.
2x3 − 6x − 2 + dx
x
x
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MATH 131-503 Fall 2015
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5.3
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3
(x2 − 3)(x + 1)dx
2.
1
Z
2
|2x − 1|dx
3.
0
Z
4. (p. 363)
1
2
(x − 1)3
dx
x2
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MATH 131-503 Fall 2015
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5.
5.3
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(6ex + ax − 2 sec2 x)dx
Net Change Theorem: The integral of a rate of change is the net change
Z b
F 0 (x)dx = F (b) − F (a)
a
This principle can be applied to all of the rates of change in the natural and social sciences that we
discussed in Section 3.8. Here are a few instances of this idea:
• If V (t) is the volume of water in a reservoir at time t, then its derivative V 0 (t) is the rate at
which water flows into the reservoir at time t. So
Z t2
V 0 (t)dt = V (t2 ) − V (t1 )
t1
is the change in the amount of water in the reservoir between time t1 and time t2 .
• If we want to calculate the distance the object travels during the time interval, we have to
consider the intervals when v(t) ≥ 0 (the particle moves to the right) and also the intervals
when v(t) ≤ 0 (the particle moves to the left). In both cases the distance is computed by
integrating |v(t)|, the speed. Therefore
Z t2
|v(t)|dt = total distance travled
t1
Example: Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per
minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30
minutes.
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