Title | 1720 review 2 |
---|---|
Author | DALLAS COWBOY |
Course | Calculus II |
Institution | University of North Texas |
Pages | 2 |
File Size | 58.9 KB |
File Type | |
Total Downloads | 57 |
Total Views | 148 |
Its the exam review ...
MATH 1720 - Practice Problems for Exam 2 Disclaimer: Many problems on the exam will resemble some of the following problems. However, you should not assume that every problem on the exam will be modeled precisely on one of these; you should also review your notes, quizzes, and homework. In problems 1 - 20 evaluate the integral. In the case of an an improper integral, determine if the integral converges, and if so, evaluate it. 1.
Z
3
xe
4x
11.
dx
Z
tan x sec x dx
12.
Z
Z
x2 + 11x dx (x − 1)(x + 1)2
13.
Z
4.
Z
dx √ 2 x x2 − 2
5.
Z
2.
3.
6.
3
Z
1 − 16x2 dx
16.
Z
e−5x sin x dx
(x2 + 4) sin 2x dx
17.
Z
3x 2 − 3x + 4 dx x(x2 + 2)
18.
Z
x3
19.
Z
2e6x
20.
Z
cos
(πx) dx
dx √ x 4 + x2 5
1
Z
1/4
8.
Z
∞
cot2 (2πx) csc6 (2πx) dx
1/8
Z
2
xe−x dx
−∞
15.
−1
Z
10.
0
Z
Z p
1
sin4 (5x) dx
14.
7.
9.
e−|x| dx
−∞
0
Z
∞
dx (x + 2)3/2
x3 + 6x2 + 17x + 16 dx x2 + 4x + 3
21. Find the partial fraction decomposition of f (x) = 22. Consider
Z
4 2
dx x−3
p
9 − 4x2 dx 6e3x dx + e3x − 3
1
x−1/2 ln x dx
0
25 . x(x + 2x + 5)2
x dx. x−1
(a) Evaluate the integral exactly. (b) Approximate the integral using the Midpoint Rule with 4 subintervals. (c) Approximate the integral using the Trapezoid Rule with 4 subintervals. (d) Approximate the integral using Simpson’s Rule with 8 subintervals. 23. Use integration by parts to derive the reduction formula; here n ≥ 2 is an integer. Z Z cosn−1 x sin x n − 1 cosn x dx = cosn−2 x dx + n n
1
Answers 1.
11e12 + 1 16
2.
sec3 x − sec x + C 3
3. 3 ln |x − 1| − 2 ln |x + 1| −
5 +C x+1
√ x2 − 2 4. +C 2x 5. x cos−1 (πx) −
√ 1 − π 2 x2 +C π
√ sin−1 (4x) x 1 − 16x2 + 6. +C 2 8 7. − 8.
x2 cos 2x x sin 2x 7 cos 2x +C − + 4 2 2
46 105π
2 9. Integral converges, √ 3 10.
x2 + 2x + 2 ln |x + 1| + 4 ln |x + 3| + C 2
11. Integral converges, 2 12.
3x sin 10x sin 20x + − +C 20 8 160
13. Integral converges, −
1 2
√ 1 4 + x2 + 2 14. − ln +C 2 x
15. Integral diverges 16. −
1 5 −5x e sin x − e−5x cos x + C 26 26
x 1 3 +C ln(x2 + 2) − √ tan−1 √ 2 2 2 !5 !3 √ √ 9 − 4x2 9 − 4x2 243 1 1 +C 18. − 3 16 5 3 3 17. 2 ln |x| +
19.
2 2 ln |e3x − 1| − ln |2e3x + 3| + C 5 5
20. Integral converges, −4 21.
1 −5x − 10 −x − 2 + + x x2 + 2x + 5 (x2 + 2x + 5)2
22. (a) 2 + ln 3; (b) ≈ 3.08975469; (c) ≈ 3.11¯6; (d) ≈ 3.09872535 23. Hint: Start with u = cosn−1 x and dv = cos x dx.
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