MATH 152 Midterm: MATH 152 TAMU 2015a X1H Solutions

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31 Jan 2019
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Mon, 23/feb c(cid:13)2015 art belmonte: use the substitution rule to nd the most general antiderivative. 5 du = dx: let u = x4 + 2. Then du = 4x3 dx or 1: hence r x3 cos(cid:0)x4 + 2(cid:1) dx = 1. 4 sin(cid:0)x4 + 2(cid:1) +c: use the substitution rule to compute the de nite integral. Show your steps by hand. (check with your calculator, if desired. ) Then du = 5 dx or 1: now u (1) = 2 and u (2) = 7, thus r 2. 2: find the area of the region enclosed by y = 5x + 30, intersections: when 5x + 30 = x3 + 2x2 24x, we have y = 6, 1, 5, area: r 5. 6(cid:12)(cid:12)(5x + 30) (cid:0)x3 + 2x2 24x(cid:1)(cid:12)(cid:12) dx. = 5581: find the area of the region bounded by x = y + 1 and x = 1.

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