BC Calculus: Review for Chapter 8 Worksheet

Evaluate the following without a calculator:

  1. \( \displaystyle \int_{0}^{2\sqrt{2}} \frac{x^3}{x^2 + 1} \, dx \)
    2. Hint long divide, then u-substitute. Should lead to a ln.
    Solution \( 4 - \ln{3} \)
  2. \( \displaystyle \int_{1}^{3} \frac{dy}{y^2 - 2y + 5} \)
    3. Hint complete the square, u-sub, then use a reference triangle. Should lead to \( \tan^{-1} \).
    Solution \( \frac{\pi}{8}\)
  3. \( \displaystyle \int_{\frac{1}{2}}^{1} \frac{\sqrt{1 - x^2}}{x^2} \, dx \)
    4. Hint use a reference triangle. Should lead to \( \cot^2 \). Change to \( csc^2 - 1 \) (Pythagorean relationship) and integrate termwise.
    Solution \( \sqrt{3} - \frac{\pi}{3}\)
  4. \( \displaystyle \int_{1}^{4} \frac{dy}{y^2 - 2y + 10} \)
    5. Hint complete the square, u-sub, and use a reference triangle. Should lead to an \( tan^{-1} \).
    Solution \( \frac{\pi}{12}\)
  5. \( \displaystyle \int_{-2}^{2} \frac{x + 2}{\sqrt{x^2 + 4x + 13}} \, dx \)
    6. Hint complete the square, u-sub, v-sub, then use the power rule.
    Solution 2
  6. \( \displaystyle \int_{0}^{5} \sqrt{25 - x^2} \, dx \)
    7. Hint referemce triangle, then use the half angle formula. (Or, if you prefer to do it without calculus, it's just a quarter circle with radius = 5.)
    Solution \( \frac{25 \pi}{4}\)
  7. \( \displaystyle \int \frac{dx}{ \left( 4 - x^2 \right) ^{\frac{3}{2}}} \)
    8. Hint use a reference triangle, reduce, and rewrite as secant sqaured. Then integrate.
    Solution \( \frac{\frac{1}{4}x}{\sqrt{4 - x^2}} + C \)
  8. \( \displaystyle \int \frac{dx}{\sqrt{x^2 + 2x}} \)
    9. Hint complete the square, u-sub, then use a reference triangle. Should lead to integral of secant.
    Solution \( \ln { \left| x + 1 + \sqrt{x^2 + 2x} \right| } + C \)
  9. \( \displaystyle \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2}{z\sqrt{4z^2 - 1}} \, dz \)
    10. Hint use a reference triangle. Should lead to inverse secant.
    Solution \( \frac{\pi}{3} \)
  10. \( \displaystyle \int_{-1}^{-\frac{1}{2}} \frac{dx}{\sqrt{-2x - x^2}} \)
    11. Hint complete the square, u-sub, then recognize it as the form of the inverse sine (or use a reference triangle.)
    Solution \( \frac{\pi}{6}\)