AP Calculus BC 6.6 Applying Properties of Definite Integrals - MCQs - Exam Style Questions
Calc-Ok Question
The graph of \(f\) consists of a semicircle centered at the origin (radius \(2\)) and three line segments. If \(\displaystyle g(x)=\int_{-2}^{x}\big(f(t)+1\big)\,dt\), find \(g(8)\).
(A) \(2\pi+6\)
(B) \(2\pi+10\)
(C) \(2\pi+15\)
(D) \(2\pi+19\)
(B) \(2\pi+10\)
(C) \(2\pi+15\)
(D) \(2\pi+19\)
▶️ Answer/Explanation
Detailed solution
\(\displaystyle g(8)=\int_{-2}^{8}\!\big(f(t)+1\big)\,dt=\int_{-2}^{8}\!f(t)\,dt+\int_{-2}^{8}\!1\,dt.\)
From the picture, \(\displaystyle \int_{-2}^{2}\!f(t)\,dt\) is the area of a semicircle of radius \(2\): \( \tfrac12\pi(2)^{2}=2\pi\).
The net signed area of the line segments on \([2,8]\) is \(+5\) (read from the graph’s geometry).
Also, \(\displaystyle \int_{-2}^{8}\!1\,dt=10\).
Hence \(g(8)=\big(2\pi+5\big)+10=2\pi+15\).
✅ Answer: (C)
Calc-OkQuestion
Let \(f\) be continuous with \(f(x)=f(-x)\) for all \(x\). If \(\displaystyle \int_{-2}^{0} f(x)\,dx=4\) and \(\displaystyle \int_{1}^{2} f(x)\,dx=1\), what is \(\displaystyle \int_{0}^{1} f(x)\,dx\)?
(A) \(-5\)
(B) \(2\)
(C) \(3\)
(D) \(5\)
(B) \(2\)
(C) \(3\)
(D) \(5\)
▶️ Answer/Explanation
Detailed solution
Even function ⇒ \(\displaystyle \int_{-a}^{0} f=\int_{0}^{a} f\).
\(\displaystyle \int_{0}^{2} f(x)\,dx=\int_{-2}^{0} f(x)\,dx=4\).
\(\displaystyle \int_{0}^{1} f=\int_{0}^{2} f-\int_{1}^{2} f=4-1=3\).
✅ Answer: (C)