AP Calculus BC 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation - MCQs - Exam Style Questions
No-CalcQuestion
If \(f'(x)=6x^2-2x+5\) and \(f(1)=-3\), then \(f(2)=\)
A. \(-19\)
B. \(8\)
C. \(13\)
D. \(22\)
B. \(8\)
C. \(13\)
D. \(22\)
▶️ Answer/Explanation
Detailed solution
\(f(x)=\int(6x^2-2x+5)\,dx=2x^3-x^2+5x+C\).
Use \(f(1)=-3\Rightarrow 2-1+5+C=-3\Rightarrow C=-9\).
Then \(f(2)=16-4+10-9=\mathbf{13}\).
✅ Correct: C
No-Calc Question
If \(y’=\dfrac{2x-1}{\sqrt{\,2x^{2}-2x+3\,}}\), then \(y=\)
(A) \(\sqrt{\,2x^{2}-2x+3\,}+C\)
(B) \(2\sqrt{\,2x^{2}-2x+3\,}+C\)
(C) \(\tfrac12\sqrt{\,2x^{2}-2x+3\,}+C\)
(D) \(\dfrac{2x^{2}-2x+3}{\sqrt{\,2x^{2}-2x+3\,}}+C\)
(B) \(2\sqrt{\,2x^{2}-2x+3\,}+C\)
(C) \(\tfrac12\sqrt{\,2x^{2}-2x+3\,}+C\)
(D) \(\dfrac{2x^{2}-2x+3}{\sqrt{\,2x^{2}-2x+3\,}}+C\)
▶️ Answer/Explanation
Detailed solution
Let \(u=2x^{2}-2x+3\Rightarrow du=(4x-2)\,dx=2(2x-1)\,dx\), so \((2x-1)\,dx=\tfrac12\,du\).
\(\displaystyle \int \frac{2x-1}{\sqrt{u}}\,dx=\tfrac12\int u^{-1/2}\,du=u^{1/2}+C=\sqrt{\,2x^{2}-2x+3\,}+C.\)
✅ Answer: (A)