A function is defined on by s(x)= cases -1, & -6 x -2, 4-x^ 2 , & -2< x 2,…

MATHEMATICS · Multi-part · CBSE · Mathematics · Integrals · Integrals

Problem

A function ss is defined on [−6,6][-6,6] by s(x)={−1,−6≤x≤−2,4−x2,−2<x≤2,0.5(x−2)−1,2<x≤6.s(x)=\begin{cases} -1, & -6\le x\le -2,\\ \sqrt{4-x^{2}}, & -2< x\le 2,\\ 0.5(x-2)-1, & 2< x\le 6. \end{cases} The graph therefore contains a horizontal segment, an upper semicircle of radius 22, and a slanted segment below the xx-axis. (1) Part 1: Compute ∫−6−2s(x) dx\displaystyle \int_{-6}^{-2} s(x)\,dx and ∫26s(x) dx\displaystyle \int_{2}^{6} s(x)\,dx, stating the geometric figures used in each calculation. (2) Part 2: Compute ∫−22s(x) dx\displaystyle \int_{-2}^{2} s(x)\,dx by interpreting the graph on that interval geometrically. (3) Part 3: Using properties of definite integrals and your answers to parts (a) and (b), determine ∫−66s(x) dx\displaystyle \int_{-6}^{6} s(x)\,dx. State whether the result is positive, negative, or zero and explain why. (4) Part 4: Let y(x)=2s(3x)y(x)=2s(3x). Without performing any new area computations, evaluate ∫−22y(x) dx\displaystyle \int_{-2}^{2} y(x)\,dx. Provide a clear explanation of how vertical and horizontal scaling factors affect the value of a definite integral.

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