AP Calculus AB 6.7 The Fundamental Theorem of Calculus and Definite Integrals - MCQs - Exam Style Questions
No-Calc Question
If \(\displaystyle \int_{1}^{k}\frac{1}{\sqrt{x}}\,dx=10\), where \(k\) is a constant, what is the value of \(k\)?
(A) \(\tfrac{1}{121}\)
(B) \(36\)
(C) \(121\)
(D) \(441\)
(B) \(36\)
(C) \(121\)
(D) \(441\)
▶️ Answer/Explanation
\[ \int \frac{1}{\sqrt{x}}\,dx=\int x^{-1/2}dx=2\sqrt{x}+C. \] Therefore \[ \int_{1}^{k}\frac{1}{\sqrt{x}}\,dx =\Big[2\sqrt{x}\Big]_{1}^{k} =2(\sqrt{k}-1)=10. \] Solve: \[ \sqrt{k}-1=5 \;\Rightarrow\; \sqrt{k}=6 \;\Rightarrow\; k=36. \] ✅ Answer: (B)
No-Calc Question
\(\displaystyle \int_{-1}^{1}\frac{x^{2}-x}{x}\,dx\) is
(A) \(-2\)
(B) \(0\)
(C) \(\tfrac{4}{3}\)
(D) nonexistent
(B) \(0\)
(C) \(\tfrac{4}{3}\)
(D) nonexistent
▶️ Answer/Explanation
Simplify for \(x\neq0\): \(\dfrac{x^{2}-x}{x}=x-1\).
Split at \(0\): \(\int_{-1}^{0}(x-1)\,dx+\int_{0}^{1}(x-1)\,dx\).
Antiderivative: \(\tfrac{x^{2}}{2}-x\).
Evaluate: \([0-0]-\left(\tfrac{1}{2}+1\right)+\left(\tfrac{1}{2}-1\right)-[0-0]=-2\).
✅ Answer: (A)