Solutions to Integration Practice Problems for AP Calculus

Practice problems for these solutions can be found at: Integration Practice Problems for AP Calculus

No calculators are allowed except for verifying your results.

1. 
2. Rewrite:
3. 
   Rewrite:
4. 
5. Let u = x – 1; du = dx and (u +1) = x.
   
6. 
7. 
8. Let u = cos x; du = –sin x dx or –du = sin x dx.
   
9. 
   Let u = x + 1; du = dx.
   
10. 
11. 
12. 
13. Since e^x and ln x are inverse functions:
    
14. 
15. Rewrite:
    
16. 
17. Since
    The point (0, 6) is on the graph of y.
    Thus, 6 = e⁰ + 2(0) + C 6 = 1 + C or C = 5. Therefore, y = e^x +2x +5.
18. Let u = e^x; du = e^xdx.
    Rewrite:
    
19. Let u = e^x + e^–x; du = (e^x – e^–x ) dx.
    
20. Since f (x) is the antiderivative of
    Given f (1) = 5; thus ln (1) + C = 5 0 + C = 5 or C = 5.
    Thus, f (x) = ln |x| + 5 and f (e) = ln (e) + 5 = 1 + 5 = 6.
21. Integrate dx by parts. Let u = x², du = 2x dx, Use parts again with u = x, du = dx, so that Integrate for and simplify to .
22. For 3x² sin x dx, use integration by parts with u = 3x², du =6x dx, dv = sin x dx, and v = – cos x. 3x² sin x dx = – 3x² cos x + [ 6x sin x – 6 sin x dx] = –3x² cos x + 6x sin x + 6 cos x + C.
23. Factor the denominator so that . Use a partial fraction decomposition, which implies Ax + A + Bx – 4B = x. Solve A + B = 1 and A – 4B to find . Integrate ln |x – 4| + ln |x +1| +C = ln |(x – 4)⁴ (x +1)| + C.
24. 
25. Begin with integration by parts, using
    
    Then
    Use partial fractions to decompose
    Solve to find
    

 

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26.  
    1. At t = 4, speed is 5 which is the greatest on 0 ≤ t ≤ 10.
    2. The particle is moving to the right when 6 < t < 10.
27. 



Since r ≥ 0,

28. Let u = ln x;

     

    Rewrite:

29. Label given points as A, B, C, D, and E.

Since f ''(x) > 0 f is concave upward for all x in the interval [0, 2].



Therefore, 1.5 < f '(1) < 2.5, choice (c). See Figure 10.8–1.



30.  
    1. f '' is decreasing on [1, 6) f '''< 0 f ' is concave downward on [1, 6) and f '' is increasing on (6, 8] f ' is concave upward on (6, 8]. Thus, at x = 6, f ' has a change of concavity. Since f '' exists at x = 6 (which implies there is a tangent to the curve of f ' at x = 6), f has a point of inflection at x = 6.
    2. f '' > 0 on [1, 4] f ' is increasing and f '' < 0 on (4, 8] f ' is decreasing. Thus at x = 4, f ' has a relative maximum at x = 4. There is no relative minimum.
    3. f '' is increasing on [6, 8] f > 0 f ' is concave upward on [6, 8].
    x = 3 cos² θ so
31. 
32. At t = 2, x = 4 sin(2π) = 0, and y = 2² – 3 · 2 + 1 = – 1, so the position of the object at t = 2 is (0, –1).
33. To find the slope of the tangent line to the curve r = 3 cos θ when θ = begin with x = r cos θ and y = r sin θ, and find and

= – 6 cos θ sin θ. y = 3 cos θ sin θ so

= 3 cos² θ – 3 sin² θ. Then the slope of the tangent line is

 



Evaluate at θ = to get The slope of the tangent line is zero, indicating that the tangent is horizontal.