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Solutions to Integration Practice Problems for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

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. Rewrite:
    • Rewrite:
  2. Let u = x – 1; du = dx and (u +1) = x.
  3. Let u = cos x; du = –sin x dx or –du = sin x dx.
    • Let u = x + 1; du = dx.
  4. Since ex and ln x are inverse functions:
  5. Rewrite:
  6. 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 = ex +2x +5.
  7. Let u = ex; du = exdx.
      Rewrite:
  8. Let u = ex + e–x; du = (exe–x ) dx.
  9. 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.
  10. 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 .
  11. 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.
  12. 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.
  13. Begin with integration by parts, using
      Then
      Use partial fractions to decompose
      Solve to find
  14.  

    ''Calculator"indicates that calculators are permitted.

  15.  
    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.
  16. Since r ≥ 0,

  17. Let u = ln x;

     

    Rewrite:

  18. Label given points as A, B, C, D, and E.
  19. 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.

    Solutions to Practice Problems

  20.  
    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].
    4. x = 3 cos2 θ so
  21. 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).
  22. 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
  23. = – 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.

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