AP Calculus BC 6.9 Integrating Using Substitution - Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Detailed solution
Use the substitution \(t=3x+4\).
Then \(dt=3\,dx \Rightarrow dx=\tfrac{1}{3}\,dt\).
When \(x=-1\), \(t=1\); when \(x=2\), \(t=10\).
\[ \int_{-1}^{2} f(3x+4)\,dx =\int_{1}^{10} f(t)\,\tfrac{1}{3}\,dt =\tfrac{1}{3}\big[F(t)\big]_{1}^{10} =\tfrac{1}{3}\big(F(10)-F(1)\big). \] ✅ Correct answer: (A)
Question
Evaluate \( \displaystyle \int_{0}^{\sqrt{\pi}/2} x\cos\!\big(x^{2}\big)\,dx \).
(A) \( 0 \)
(B) \( \dfrac{1}{2} \)
(C) \( \dfrac{\sqrt{2}}{4} \)
(D) \( \dfrac{\sqrt{2}}{2} \)
(B) \( \dfrac{1}{2} \)
(C) \( \dfrac{\sqrt{2}}{4} \)
(D) \( \dfrac{\sqrt{2}}{2} \)
▶️ Answer/Explanation
Detailed solution
Let \( u=x^{2} \Rightarrow du=2x\,dx \Rightarrow x\,dx=\tfrac{1}{2}du \).
Bounds: \( x=0\Rightarrow u=0 \); \( x=\tfrac{\sqrt{\pi}}{2}\Rightarrow u=\tfrac{\pi}{4} \). \[ \int_{0}^{\sqrt{\pi}/2} x\cos(x^{2})\,dx =\tfrac{1}{2}\!\int_{0}^{\pi/4}\!\cos u\,du =\tfrac{1}{2}\big[\sin u\big]_{0}^{\pi/4} =\tfrac{1}{2}\cdot\frac{\sqrt{2}}{2} =\frac{\sqrt{2}}{4}. \] ✅ Answer: (C)