BC Calculus : §6.2 U-Sub Worksheet

For each of the following, rewrite using mathematical notation, then evaluate without the use of a calculator.

  1. \( \displaystyle \int_{0}^{3} (4 - t)^{-\frac{1}{2}} \, dt \)
    2. Solution \(u = 4 - t\), answer: 2
  2. \( \displaystyle \int_{-1}^{0}(3u + 4)^{\frac{1}{2}} \, du\)
    3. Solution \(v = 3u + 4\), answer: \(\frac{2}{9} \left( 4^{\frac{3}{2}} - 1 \right) \)
  3. \( \displaystyle \int_{2}^{3}(2y - 3)^{-1} \, dy\)
    4. Solution \(u = 2y - 3\), answer: \(\frac{\ln{3}} {2} \)
  4. \( \displaystyle \int_{0}^{\sqrt{3}} x(4 - x^2)^{-\frac{1}{2}} \, dx\)
    5. Solution \(u = 4 - x^2\), answer: 1
  5. \( \displaystyle \int_{1}^{2} \frac{3x - 1}{3x} \, dx\)
    6. Solution Break it up into two integrals. Answer: \(1 - \frac{\ln{2}}{3}\)
  6. \( \displaystyle \int_{0}^{1} e^{-x} \, dx\)
    7. Solution \(u = -x\), answer: \(1 - \frac{1}{e}\)
  7. \( \displaystyle \int_{0}^{1} xe^{x^2} \, dx\)
    8. Solution \(u = x^2\), answer: \(\frac{e}{2} - \frac{1}{2}\)
  8. \( \displaystyle \int_{1}^{e} \frac{\ln{x}}{x} \, dx\)
    9. Solution \(u = \ln{x}\), answer: \(\frac{1}{2}\)
  9. \( \displaystyle \int_{0}^{\frac{\pi}{6}} \frac{ \cos{t}}{1 + 2 \sin{t}} \, dt\)
    10. Solution \( u = 1 + 2 \sin{t}\), answer: \( \frac{\ln{2}}{2}\)
  10. \( \displaystyle \int_{0}^{1} \frac{e^{-x} +1}{e^{-x}} \, dx\)
    11. Solution \( \displaystyle u = e^{-x}\) then... grrr. I fell into my own trap. Break it up! Answer: \(e\)