Directions: Show all work to receive full credit.
Name:
- Solve the following integrals.
(a) R
x
(2x+1)3 dx
1
4
(
1
2(2x+1)2 −
1
2x+1 )
(b) R
x cos2
(x) dx
x
2
(x +
sin(2x)
2
) −
1
2
(
x
2
2 −
cos(2x)
4
)
(c) R ln(√
x)
x
dx
1
4
(ln(x))2
(d) R
x sin(x) cos(x) dx
1
2
(−
x
2
cos(2x) + 1
4
sin(2x))
Calculus II Exam 02 Review Page 2 of 5
(e) R
1
(x2−1)2 dx
− ln | √ x
x2−1
- √
1
x2−1
|
(f) R
x
√
x − 3 dx
2
5
(x − 3)5/2 + 2(x − 3)3/2 + C
(g) R
x
2
ln(x) dx
x
3
3
ln(x) −
1
9
x
3 + C
(h) R √
1+x2
x
dx
ln |
√
1+x2
x −
1
x
| +
√
x
2 + 1 + C
Cont.
Calculus II Exam 02 Review Page 3 of 5
(i) R
e
x
1+e
x dx
ln(e
x+1) Think about why there is no absolute value bars
(j) R
sin2
(x) cos2
(x) dx
1
8
(x −
sin(4x)
4
) + C
(k) R
x √
2+1
x2+2x+2 dx (Note: This took me almost an entire page. Not Mandatory.)
(
√
(x+1)2+1)(x+1)+3 ln |
√
(x2+1)2+1+(x+1)|
2 −2
p
(x + 1)2 + 1
(l) R
x−4
x2−5x+6 dx
− ln |x − 3| + 2 ln |x − 2| + C
Cont.
Calculus II Exam 02 Review Page 4 of 5
(m) R
tan2
(x) sec4
(x) dx
tan5
(x)
5 +
tan3
(x)
3 + C
(n) R ln(√
x)
√
x
dx
2(√
x ln(√
x) −
√
x) + C
(o) R
arctan(x) dx
x arctan(x)−
1
2
ln(1+x
2
)+C Why not absolute value?
(p) R
sin3
(x) cos2
(x) dx
−(
cos3
(x)
3 −
cos5
(x)
5
) + C
Cont.
Calculus II Exam 02 Review Page 5 of 5
- Use Simpson’s rule to estimate the integral. R 2
0
x
2
x2+1 dx where n = 8
Approximately: .8928598757 - Use the Trapazoid rule to estimate the integral. R 3
1
sin(x)
x
dx where n = 4
Approximately: .90164486 - Determine the number of subintervals are needed to estimate the integral R 2
0
xex dx accurate to within
0.001 units using Simpson’s rule.
n ≥ 10
The End.
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