Consider the function q(x)= cases 1 (x-1)^ 2 , & 0 x<1, 2-x, & 1< x 2

MATHEMATICS · Single correct · AP · Calculus BC · Integration and Accumulation of Change · Integration and Accumulation of Change

Problem

Consider the function q(x)={1(x−1)2,0≤x<1,2−x,1<x≤2.q(x)=\begin{cases} \dfrac{1}{(x-1)^{2}}, & 0\le x<1,\\ 2-x, & 1< x\le 2. \end{cases} Which of the following statements is true concerning the area between the graph of qq and the xx–axis on 0≤x≤20\le x\le 2? (a) The area is finite and equals ∫011(x−1)2 dx+∫12(2−x) dx\displaystyle\int_{0}^{1}\frac{1}{(x-1)^{2}}\,dx+\int_{1}^{2}(2-x)\,dx. (b) The area is infinite because both ∫011(x−1)2 dx\displaystyle\int_{0}^{1}\frac{1}{(x-1)^{2}}\,dx and ∫12(2−x) dx\displaystyle\int_{1}^{2}(2-x)\,dx diverge. (c) The area is infinite because ∫011(x−1)2 dx\displaystyle\int_{0}^{1}\frac{1}{(x-1)^{2}}\,dx diverges, whereas ∫12(2−x) dx\displaystyle\int_{1}^{2}(2-x)\,dx is finite. (d) The area is finite and equals 32\dfrac{3}{2}.

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