MATH 172 Midterm: MATH 172 TAMU 172-Spring 18 Exam 1 2 Solutions
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6x2 x3 + 1 dx. (a) 2 ln(6) (b) 2 ln(7) (c) 2 ln(8) (d) 2 ln(9) (e) 2 ln(10) Let u = x3 + 1, so du = 3x2 dx. 6x2 x3 + 1 dx = z u(2) u(0) = 2 ln 9 2 ln 1 = 2 ln 9 . Consider the region in the rst quadrant bounded by x = y2, y = 3, and the y-axis. 0 (a) z 9 (b) z 9 (c) 2 z 9 (d) 2 z 3 (e) 2 z 3. 0 (9 x) dx x2 dx x(3 x) dx y(3 y2) dy y4 dy. Solution: a we rst nd the intersection point between x = y2 and y = 3, which is easily seen to be (x, y) = (9, 3). The region in question then sits above the interval. 0 x 9 and between the curves y = x and y = 3.
