A. Armstrong
MA 241- Summer 2014 - Integration Review
Indefinite Integrals
•
•
•
•
•
•
•
R
R
R
[f (x) + g(x)]dx = f (x)dx + g(x)dx
R
R
cf (x)dx = c f (x)dx
Z
xn+1
xn dx =
+ C, (n 6= −1)
n+1
Z
1
dx = ln | x | +C
x
R x
x
C
Z e dx = e +
x
a
ax dx =
+C
ln a
R
sin xdx = − cos x + C
•
R
cos xdx = sin x + C
•
R
sec2 xdx = tan x + C
•
R
csc2 xdx = − cot x + C
•
R
sec x tan xdx = sec x + C
R
• Z csc x cot xdx = − csc x + C
1
√
dx = sin−1 x + C
•
2
1−x
Z
1
•
dx = tan−1 x + C
x+1
Integration Techniques
• u-substitution: If u = g(x) is differentiable and f (x) is continuous, then
R
f (u)du.
R
f (g(x))g 0 (x)dx =
(1) Determine u = g(x).
(2) Find the derivative:
du
dx .
(3) Solve for dx.
(4) Substitute back into the integral. Simplify integral so the only variable is u.
(5) Integrate with respect to u.
(6) Substitute back for x. If a definite integral, solve as usual.
R
R
R
• Integration by Parts: Given f (x)g 0 (x)dx, let u = f (x) and v = g(x). Then udv = uv − vdu
where du = f 0 (x)dx and dv = g 0 (x)dx.
1
Review Problems
1. Evaluate
the following integrals.
Z
3
(a) (1 + 8x − x2 )dx
7
Z
4 + x2
(b)
dx
3
Z 2 x
2ex + 4 cos xdx
(c)
Z0 3/2
6
√
dt
(d)
1 − t2
1/2
2. Use substitution
to evaluate the following integrals.
Z
2
(a)
xex dx
Z
x
(b)
dx
2
Z x +1
(c)
esin x cos xdx
Z
(d)
x2 (1 + 2x3 )5 dx
Z
3. Use integration by parts twice to find the following integral:
Z
4. Evaluate the following integral using integration by parts:
x2 sin xdx
1
(x2 + 1)ex dx
0
5. Evaluate the following integrals. (Note: You may need to use u-substitution or integration by
partsZto solve the integral.)
(a)
x3 (1 − x4 )8 dx
Z
(b)
t(1 − t)2 dt
Z
(c)
sec x(sec x + tan x)dx
Z
(d)
x5 ln(x)dx
Z p
5
(e)
9 − 5y dy