AP Calculus AB 6.9 Integrating Using Substitution - MCQs - Exam Style Questions
Calc-Ok Question
The function \(f\) is continuous for all \(x\), and \(\displaystyle \int_{1}^{2} f(3x-2)\,dx=a\). Which is equal to \(\displaystyle \int_{1}^{4} f(u)\,du\)?
(A) \(\tfrac{a}{3}\)
(B) \(a\)
(C) \(2a\)
(D) \(3a\)
▶️ Answer/Explanation
Substitute \(u=3x-2\Rightarrow du=3\,dx\Rightarrow dx=\tfrac{1}{3}du\).
When \(x=1\), \(u=1\); when \(x=2\), \(u=4\).
So \(a=\displaystyle\int_{1}^{2} f(3x-2)\,dx=\tfrac{1}{3}\int_{1}^{4} f(u)\,du\).
Hence \(\displaystyle \int_{1}^{4} f(u)\,du=3a\).
✅ Answer: (D)
When \(x=1\), \(u=1\); when \(x=2\), \(u=4\).
So \(a=\displaystyle\int_{1}^{2} f(3x-2)\,dx=\tfrac{1}{3}\int_{1}^{4} f(u)\,du\).
Hence \(\displaystyle \int_{1}^{4} f(u)\,du=3a\).
✅ Answer: (D)
No-Calc Question
\[ \int\!\left(\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt[3]{\,1-x\,}}\right)\,dx \;=\; ? \]
(A) \( \dfrac{x}{(1-x^{2})^{3/2}}+\dfrac{1}{3(1-x)^{4/3}}+C\)
(B) \( 2\sqrt{1-x^{2}}+\dfrac{3}{2}(1-x)^{2/3}+C\)
(C) \( \sin^{-1}x+\dfrac{2}{3}(1-x)^{2/3}+C\)
(D) \( \sin^{-1}x-\dfrac{3}{2}(1-x)^{2/3}+C\)
▶️ Answer/Explanation
Use known antiderivative: \(\displaystyle \int \frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x+C\).
For \(\displaystyle \int (1-x)^{-1/3}dx\): let \(u=1-x\), \(du=-dx\).
Then \( \int (1-x)^{-1/3}dx = -\int u^{-1/3}du = -\frac{u^{2/3}}{2/3} = -\frac{3}{2}(1-x)^{2/3}\).
Sum the parts:
\( \sin^{-1}x – \dfrac{3}{2}(1-x)^{2/3}+C\).
✅ Answer: (D)