ReviewExam1Solutions

ReviewExam1Solutions - Math136 Solutions to Review for exam...

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Math136 Solutions to Review for exam 1 Summer 2010 1. Suppose R is the region in the first quadrant bounded by the curves y = 1 x , y = x 2 27 and x = 1. Set up and evaluate an integral whose value is (a) the area of R . The curves 1 /x and x 2 / 27 intersect at x = 3. The area of R is given by A = Z 3 1 1 x - x 2 27 dx = ± ln x - x 3 81 ²³ ³ ³ ³ 3 1 = ln3 - 1 3 - (ln1 - 1 81 ) = ln3 - 26 81 (b) the volume of the solid resulting when revolving R about the x - axis. Using the washer method, f ( x ) = 1 /x , g ( x ) = x 2 / 27 so the volume is V = π Z 3 1 ± 1 x ² 2 - ± x 2 27 ² 2 dx = π Z 3 1 1 x 2 - x 4 729 dx = π ± - 1 x - x 5 3645 ²³ ³ ³ ³ 3 1 = π ´± - 1 3 - 1 15 ² - ± - 1 - 1 3645 ²µ = 2188 3645 π (c) the volume of the solid resulting when revolving R about the y - axis. Using the shell method, r = x and h = 1 /x - x 2 / 27, so the volume is V = 2 π Z 3 1 x ± 1 x - x 2 27 ² dx = 2 π Z 3 1 ± 1 - x 3 27 ² dx = 2 π ± x - x 4 108 ²³ ³ ³ ³ 3 1 = 2 π ´± 3 - 81 108 ² - ± 1 - 1 108 ²µ = 68 27 π 2. Find the volume of the solid S obtained by revolving the region enclosed by x = 0, x = 1, y = 0, and y = e x , about the y - axis. Using the shell method, r = x , h = e x so the volume is V = 2 π Z 1 0 xe x dx We use integration by parts with u = x , dv = e x , so du = dx and v = e x . Then V = 2 π ´ xe x | 1 0 - Z 1 0 e x dx µ = 2 π [( e - 0) - e x | 1 0 ] = 2 π [ e - ( e - 1)] = 2 π
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3. Compute the future value of an account earning interest at a rate of 7% 5 years from now if money is deposited at a rate of f ( t ) = 10 + 3 t dollars per year. FV = Z 5 0 (10 + 3 t ) e . 07(5 - t ) dt We use integration by parts with u = 10 + 3 t , dv = e . 07(5 - t ) dt , so du = 3 dt and v = 1 - . 07 e . 07(5 - t ) . Then FV = (10 + 3 t ) 1 - . 07 e . 07(5 - t ) ± ± ± ± 5 0 - Z 5 0 1 - . 07 e . 07(5 - t ) 3 dt = ² 25 - . 07 - 10 - . 07 e . 35 ³ - 3 ( - . 07) 2 e . 07(5 - t ) ± ± ± ± 5 0 = ² 25 - . 07 - 10 - . 07 e . 35 ³ - 3 ( - . 07) 2 (1 - e . 35 ) = 102 . 15 4. The base of a solid S is the region on the xy plane enclosed by the semicircle above the x - axis x 2 + y 2 = 4. Its cross sections perpendicular to the y - axis are rectangles with height half the base. Find the volume of the solid S . For any value
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This note was uploaded on 10/23/2011 for the course MATH 136 taught by Professor Prellis during the Spring '08 term at Rutgers.

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ReviewExam1Solutions - Math136 Solutions to Review for exam...

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