BC Calculus Mixed Integration Practice
- Class: 5H: BC Calculus
- Author: Peter Atlas
- Text: Calculus Finney, Demana, Waits, Kennedy
The problems in this worksheet may or may not involve techniques learned in chapter 8. Calculator Inactive.
- \( \displaystyle \int sin^{-1}(x) \, dx = \)
- \( \displaystyle \left( 1 - x^2 \right) ^{-\frac{1}{2}} + C \)
- \( \displaystyle x\sin^{-1}{x} + \left( 1 - x^2 \right) ^{\frac{1}{2}} + C \)
- \( \displaystyle x\sin^{-1}{x} + \frac{1}{2} \left( 1 - x^2 \right) ^{\frac{1}{2}} + C \)
- \( \displaystyle x\sin^{-1}{x} - \left( 1 - x^2 \right) ^{\frac{1}{2}} + C \)
- \( \displaystyle x\sin^{-1}{x} + \frac{1}{2} \ln { \left| 1 - x^2 \right|} + C \)
Solution
B
- \( \displaystyle \int_2^3 \frac{dx}{x^2 - 1} = \)
- \( \displaystyle \frac{1}{2} \ln { \frac{2}{3}} \)
- \( \displaystyle \frac{1}{2} \ln { \frac{3}{2} } \)
- \( \displaystyle \ln { \frac{3}{2}} \)
- \( \displaystyle \ln { \frac{2}{3} } \)
- \( \displaystyle 2 \ln { \frac{3}{2}} \)
Solution
B
- Find the area bounded by \( \displaystyle f(x) = \frac{1}{x(2 - x)}\) and the x-axis on the interval [3, 5].
- \( \displaystyle \frac{1}{2} \ln { \frac{5}{9}} \)
- \( \displaystyle \ln { \frac{5}{9}} \)
- \( \displaystyle \ln { \frac{3}{\sqrt{5}} } \)
- \( \displaystyle \ln { \frac{9}{5} } \)
- \( \displaystyle \frac{1}{2} \ln {12} \)
Solution
C
- \( \displaystyle \int_4^6 \frac{x - 1}{x^2- 4} \, dx =\)
- \( \displaystyle \frac{7 \ln {2} + 3 \ln {3}}{4} \)
- \( \displaystyle \frac{7 \ln {2} - 3 \ln {3}}{4} \)
- \( \displaystyle \frac{7}{4} \ln {\frac{3}{2} } \)
- \( \displaystyle \frac{4}{7} \ln { \frac{3}{2} } \)
- \( \displaystyle \frac{7 \ln {2} - 2 \ln {3}}{4} \)
Solution
B
- \( \displaystyle \int x \cos{x^2} \, dx = \)
- \( \displaystyle x \sin{x^2} + C \)
- \( \displaystyle \frac{1}{2} \sin{x^2} + C \)
- \( \displaystyle -\frac{1}{2} \sin{x^2} + C \)
- \( \displaystyle \cos{x^2} + C \)
- \( \displaystyle \sin{x^2} + C \)
Solution
B
- \( \displaystyle \int_1^{e^2} x^2 \ln {x^2} \, dx = \)
- \( \displaystyle \frac{e^6 - 2}{9} \)
- \( \displaystyle \frac{e^6 + 2}{9} \)
- \( \displaystyle \frac{10e^6 - 2}{9} \)
- \( \displaystyle \frac{10e^6 + 2}{9} \)
- \( \displaystyle \frac{10e^6 + 1}{9} \)
Solution
D
- \( \displaystyle \int \frac{x^2 + 5x}{x^2 + 5x + 6} \, dx = \)
- \( \displaystyle -6 \ln {\left| \frac{x + 2}{x + 3} \right| } + C \)
- \( \displaystyle x - 6 \ln { \left| \frac{ x + 2}{x + 3} \right| } + C \)
- \( \displaystyle -5 \ln { \left| \frac{ x + 2}{x + 3} \right| } + C \)
- \( \displaystyle - \ln { \left| \frac{ x + 2}{x + 3} \right| } + C \)
- \( \displaystyle \frac{x}{6} + C \)
Solution
B
- \( \displaystyle \int \frac{(x + 2)^3}{x^2 + 2x} \, dx =\)
- \( \displaystyle 4 \ln {|x|} + C \)
- \( \displaystyle \frac{1}{2} x^2 + 4x + \ln {|x|} + C \)
- \( \displaystyle \frac{1}{2} x^2 + 2x + 4 \ln {|x|} + C \)
- \( \displaystyle \frac{1}{2} x^2 + 4x + 4 \ln {|x|} + C \)
- \( \displaystyle \frac{1}{2} x^2 + 4 \ln {|x|} + C \)
Solution
D
- \( \displaystyle \int_1^{\infty} \frac{2}{\sqrt[3]{x}} \, dx = \)
- \( \displaystyle -\infty \)
- \( \displaystyle - \frac{2}{3} \)
- 0
- \( \displaystyle \frac{2}{3} \)
- \( \displaystyle \infty \)
Solution
E
- \( \displaystyle \int_0^1 \frac{dx}{\sqrt[3]{x - 1}} = \)
- \( \displaystyle -\infty \)
- \( \displaystyle -\frac{3}{2} \)
- \( \displaystyle 0 \)
- \( \displaystyle \frac{3}{2} \)
- \( \displaystyle \infty \)
Solution
B
- \( \displaystyle \int_1^r \frac{dx}{x \ln{x}}\) where \(r\) is a real number greater than 1 =
- 0
- 1
- \( \displaystyle \ln {r} \)
- \( \displaystyle -\infty \)
- \( \displaystyle \infty \)
Solution
E
- When \(n\) is an integer greater than 1, \( \displaystyle \int_0^1 \frac{dx}{x^n} = \)
- 0
- \( \displaystyle \frac{1}{-n + 1} \)
- 1
- \( \displaystyle -\infty \)
- \( \displaystyle \infty \)
Solution
E
- When \(n\) is an integer greater than 1, \( \displaystyle \int_1^{\infty} \frac{dx}{x^n} = \)
- 0
- 1
- \( \displaystyle \frac{1}{n - 1} \)
- \( \displaystyle -\infty \)
- \( \displaystyle \infty \)
Solution
C
- \( \displaystyle \int_1^3 x^3 \ln {x } \, dx = \)
- \( \displaystyle 27 \ln { 3} - 12 \)
- \( \displaystyle 27 \ln { 3} - \frac{27}{2} \)
- \( \displaystyle \frac{81}{4} \ln { 3} - \frac{81}{16} \)
- \( \displaystyle \frac{81}{4} \ln { 3} - 5 \)
- \( \displaystyle 27 \ln {3} \)
Solution
D
- \( \displaystyle \int_0^{\frac{\pi}{4}} x \sec^2 {x} \, dx =\)
- \( \displaystyle \frac{\pi}{4} \)
- \( \displaystyle \frac{\pi}{4} + 1 \)
- \( \displaystyle \frac{\pi}{4} - \frac{3}{2} \ln {2} \)
- \( \displaystyle \frac{\pi}{4} - \frac{1}{2} \ln {2} \)
Solution
D
- \( \displaystyle \int_1^2 \frac{x^3 dx}{\sqrt{x^2 - 1}} = \)
- 0
- \( \displaystyle \sqrt{3} \)
- \( \displaystyle 2\sqrt{3} \)
- \( \displaystyle 4\sqrt{3} \)
- \( \displaystyle \infty \)
Solution
C
- \( \displaystyle \int_1^2 \frac{x^4 + 1}{x^3 + x^2} \, dx =\)
- \( \displaystyle \ln {\frac{9}{8} + 1} \)
- \( \displaystyle \ln {\frac{8}{9} + 1} \)
- \( \displaystyle \ln {\frac{9}{8}} + \frac{1}{2} \)
- \( \displaystyle \ln {\frac{8}{9}} + \frac{1}{2} \)
- \( \displaystyle \ln {\frac{9}{8}} \)
Solution
A
- Find the area under the curve \( \displaystyle f(x) = \frac{1}{ x^2 - x}\) for \(x \geq 2\).
- \( \displaystyle - \ln {2} \)
- \( \displaystyle \ln {2} \)
- 1.25
- \( \displaystyle e^2 \)
- \( \displaystyle \infty \)
Solution
B
- Find \( \displaystyle \lim_{n \to \infty} \int_0^2 \frac{dx}{\sqrt[n]{x}}\) for \(n > 1\).
- 0
- 1
- 2
- \( \displaystyle 2\sqrt{2} \)
- \( \displaystyle \infty \)
Solution
C
- The region bounded by the graph of \( \displaystyle y = \tan {x}\), the x-axis, and the line \( \displaystyle x = \frac{\pi}{4}\) is the base of a solid. Find the volume of the solid if cross-sections perpendicular to the x-axis are squares.
- 1
- \( \displaystyle 1 - \frac{\pi}{4} \)
- \( \displaystyle 4 - \pi \)
- \( \displaystyle 4 - \frac{\pi}{4} \)
- \( \displaystyle \sqrt{3} - \frac{\pi }{3} \)
Solution
B
- If \( \displaystyle f(x) = \frac{1}{\sqrt{x^2 - 1}} \), then \( \displaystyle \int_1^2 xf(x) \, dx =\)
- \( \displaystyle \sqrt{3} \)
- \( \displaystyle \sqrt{3} - 1 \)
- \( \displaystyle 1/\sqrt{3} \)
- 1
- \( \displaystyle \infty \)
Solution
A
- \( \displaystyle \int x^3 f''\left( x^2 \right) \, dx = \)
- \( \displaystyle \frac{1}{2} \left( x^2f(x^2) - f(x^2) \right) + C \)
- \( \displaystyle \frac{1}{2} \left( x^2f'(x^2) - f(x^2) \right) + C \)
- \( \displaystyle x^2f'(x^2) - f(x^2) + C \)
- \( \displaystyle 2 \left( x^2f'(x^2) - f(x^2) \right) + C \)
- \( \displaystyle \frac{1}{2} \left( x^2f''(x^2) - f'(x^2) \right) + C \)
Solution
B
- Find the area between the x-axis and the curve \( \displaystyle f(x) = \frac{1}{ x^2 + x - 6}\) on the interval [-2, 0].
- \( \displaystyle \ln { \frac{3}{2}} \)
- \( \displaystyle \frac{1}{5} \ln { \frac{3}{2}} \)
- \( \displaystyle \frac{1}{5} \ln {6} \)
- \( \displaystyle \ln {6} \)
- undefined
Solution
C
- \( \displaystyle f(x) = 2x^2, g(x) = \csc{x}\). Find \( \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} f \left( g(x) \right) \, dx. \)
- -2
- -1
- 1
- \( \displaystyle \sqrt{3} \)
- 2
Solution
E
- \( \displaystyle \int \frac{dx}{5 + 5 \cos{x}} = \)
- \( \displaystyle 5x + 5 \sin{x} + C \)
- \( \displaystyle \frac{1}{5x + 5 \sin{x} } + C \)
- \( \displaystyle \frac{1}{5} \tan {\frac{x}{2}} + C \)
- \( \displaystyle \ln {(5 + 5\cos{x})} + C \)
- \( \displaystyle \frac{1}{5} \left( x + \ln {\left| \sec{x} + \tan{x} \right|} \right) +C \)
Solution
C
- \( \displaystyle \int_1^{\infty} \frac{ dx}{x \left( \ln {x} + 1 \right)^2} =\)
- \( \displaystyle \frac{1}{2} \)
- 1
- 2
- 3
- \( \displaystyle \infty \)
Solution
B
- \( \displaystyle \int_0^2 \sqrt[5]{1 - x} \, dx \) =
- 0
- \( \displaystyle \frac{5}{6} \)
- \( \displaystyle \frac{5}{4} \)
- \( \displaystyle \frac{5}{2} \)
- undefined
Solution
A
- Find the y-intercept of the line tangent to the graph of \( \displaystyle xy^2 + \ln { x} = y + 6\) at the point (1, 3).
- \( \displaystyle -\frac{23}{5} \)
- -2
- 1
- \( \displaystyle \frac{7}{5} \)
- 5
Solution
E