# Document 10504201

```c
Math 152, Benjamin
Aurispa
Review of 6.4: The Fundamental Theorem of Calculus
First, you need to make sure you refresh yourself on the antiderivatives below, since they are the foundation
of ALL integrals!!
Table of Some Antiderivatives:
Function
General Antiderivative
a (a constant)
ax + C
xn , n 6= −1
xn+1
+C
n+1
1
x
ex
ln |x| + C
ex + C
ax
+C
ln a
ax
Function
1
√
1 − x2
1
1 + x2
General Antiderivative
sin x
cos x
− cos x + C
sin x + C
sec2 x
tan x + C
sec x tan x
csc x cot x
csc2 x
sec x + C
− csc x + C
− cot x + C
arcsin x + C
arctan x + C
Recall that an indefinite integral is used to indicate the process of finding the most general antiderivative
of f (x):
Z
f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x) (i.e. F ′ (x) = f (x)).
Examples: Calculate the following general indefinite integrals.
Z √
1
3
•
4 + 7x5 + x7 − 2 dx
6x
•
Z 1
5
+ sec x tan x − √
2
1+x
x
The definite integral
interval [a, b].
A
B
Rb
a
dx
f (x) dx represents the NET AREA bounded between f (x) and the x−axis on the
C
1
c
Math 152, Benjamin
Aurispa
To evaluate definite integrals, we use the following:
The Fundamental Theorem of Calculus, Part 2:
If f is continuous on [a, b], then
Z
b
a
f (x) dx = F (b) − F (a)
where F is any antiderivative of f .
Examples: Evaluate the following definite integrals.
•
Z
•
Z
•
Z
•
Z
1
3
2
0
1
3
−
3
x
x
dx
(3ex − 2x ) dx
π
4
(sec2 θ + 2 sin θ) dθ
0
0
(t2 + 1)2 dt
−2
2
c
Math 152, Benjamin
Aurispa
•
Z
4
1
u2 − 1
√ du
u u

 √ 1
f (x) dx where f (x) =
•
1 − x2

−1/2
Z
π/2
cos x
− 21 ≤ x &lt; 0
0≤x≤
π
2
Set up the integral(s) to find the area under the curve f (x) = |x2 − 4| between x = 0 and x = 4.
For a differentiable function f , if f (8) = 12 and
Z
8
f ′ (x) dx = 9, what is f (−1)?
−1
3
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Math 152, Benjamin
Aurispa
6.5: Substitution
Recall the Chain Rule for derivatives:
d
[g(x)]n = n[g(x)]n−1 g′ (x)
dx
d
f (g(x)) = f ′ (g(x))g ′ (x)
dx
d
sin(g(x)) = cos(g(x))g ′ (x)
dx
d g(x)
e
= eg(x) g′ (x)
dx
d
g′ (x)
ln(g(x)) =
dx
g(x)
The Substitution Rule for integrals is essentially the reverse of the Chain Rule: If u = g(x) is a differentiable function, then
Z
Z
f (g(x))g ′ (x) dx = f (u) du
Examples:
Z
3x2 (2 + x3 )10 dx
Z
te1−4t dt
Z
√
Z
Z
2
sin x
dx
1 + cos x
(3 sec2 x + 6) cos(tan x + 2x) dx
e5z + 2
dz
e5z + 10z
4
c
Math 152, Benjamin
Aurispa
Z
1
dx
x ln(x)
Z
arcsin 4x
√
dx
1 − 16x2
Z
tan x dx
Z
e3x dx
Z
sin(11x − 5) dx
Z
√
3
1
dx
9−x
5
c
Math 152, Benjamin
Aurispa
Z
5 + 7x
dx
x2 + 1
Z
x(x + 9)7 dx
Z
√
x3
dx
5x2 − 1
6
c
Math 152, Benjamin
Aurispa
For definite integrals, there are two ways to evaluate when using substitution:
Z
1
Z
9
e x
√ dx
x
Z
4
x
dx
6−x
0
4
1
p
x2 x3 + 9 dx
√
7
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