AP Calculus BC 6.11 Integrating Using Integration by Parts - MCQs - Exam Style Questions
▶️ Answer/Explanation
Choose integration by parts twice.
First IBP:
Let \(u=x^{2}\), \(dv=e^{2x}dx\) \(\Rightarrow\) \(du=2x\,dx\), \(v=\tfrac{1}{2}e^{2x}\).
\(\displaystyle \int x^{2}e^{2x}dx = \tfrac{1}{2}x^{2}e^{2x} – \int x\,e^{2x}dx.\)
Second IBP on \(\displaystyle \int x\,e^{2x}dx\):
Let \(u=x\), \(dv=e^{2x}dx\) \(\Rightarrow\) \(du=dx\), \(v=\tfrac{1}{2}e^{2x}\).
\(\displaystyle \int x\,e^{2x}dx = \tfrac{1}{2}x\,e^{2x} – \int \tfrac{1}{2}e^{2x}dx = \tfrac{1}{2}x\,e^{2x} – \tfrac{1}{4}e^{2x}+C.\)
Substitute back:
\(\displaystyle \int x^{2}e^{2x}dx = \tfrac{1}{2}x^{2}e^{2x} – \Big(\tfrac{1}{2}x\,e^{2x} – \tfrac{1}{4}e^{2x}\Big) = \tfrac{1}{2}x^{2}e^{2x} – \tfrac{1}{2}x\,e^{2x} + \tfrac{1}{4}e^{2x} + C.\)
✅ Answer: (A)
No-Calc Question
(B) \(15\)
(C) \(21\)
(D) \(27\)
▶️ Answer/Explanation
Use integration by parts.
Let \(u=x\), \(dv=g”(x)dx\) \(\Rightarrow\) \(du=dx\), \(v=g'(x)\).
\(\displaystyle \int_{0}^{3} x\,g”(x)\,dx = \Big[x\,g'(x)\Big]_{0}^{3} – \int_{0}^{3} g'(x)\,dx.\)
Evaluate with the given data:
\([x\,g'(x)]_{0}^{3}=3\times g'(3)=3\times 6\), and \(\displaystyle \int_{0}^{3} g'(x)\,dx=g(3)-g(0)=5-2.\)
Therefore \( \displaystyle \int_{0}^{3} x\,g”(x)\,dx = 18 – (5-2) = 15.\)
✅ Answer: (B)