Sketch the region enclosed by the curves y = x + 2, y = 16 - x2, x = -2, and x =

Question

Sketch the region enclosed by the curves y = x + 2, y = 16 - x2, x = -2, and x = 2. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

Explanation / Answer

Since 16-x^2>x+2 on the interval form -2 to 2 we use a difference in integrals and will integrate to find area as follows. INTEGRAL[(16-x^2)-(x+2) dx,a=-2,b=2] or (-x^2-x+14,-2 to 2). So we get [-x^3/3-x^2/2+14x]{a=-2,b=2} Plugging in bounds we get -8/3-2+28+(-8/3)+(2)-(14*-2) or (50+2/3) or 152/3 which is the area between the two graphs.

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