finalpracticesolutions - Math 2260 Final Exam Practice...

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Math 2260 Final Exam Practice Problem Solutions 1. Write down an integral which will compute the length of the part of y = ln x for 1 x 2. Don’t worry about evaluating this integral. Answer: Since dy dx = 1 x , the arc length integral is Z 2 1 s 1 + 1 x 2 dx = Z 2 1 r 1 + 1 x 2 dx = Z 2 1 r x 2 + 1 x 2 dx = Z 2 1 1 + x 2 x dx 2. Let V ( a ) be the volume obtained by rotating the area between the x -axis and the graph of y = 1 x 3 / 2 from x = a to x = 1 around the y -axis. What happens as a goes to zero? Answer: Using the cylindrical shell method, V ( a ) is given by V ( a ) = Z 1 a 2 πx 1 x 3 / 2 dx = 2 π Z 1 a 1 x 1 / 2 dx = 2 π 2 x 1 a = 2 π 2 - 2 a = 4 π (1 - a ) . Therefore, as a 0, the quantity V ( a ) 4 π . 3. What is the volume of the solid obtained by rotating the part of the graph of y = cos x sin x between x = 0 and x = π/ 2 around the x -axis? Answer: Using the disk method, the volume will be Z π/ 2 0 π cos x sin x 2 dx = Z π/ 2 0 π cos 2 x sin x dx. Then, letting u = cos x , we have that du = - sin x , and so Z π/ 2 0 π cos 2 x sin x dx = - π Z π/ 2 0 cos 2 x ( - sin x ) dx = - π Z 0 1 u 2 du = - π u 3 3 0 1 = - π 0 - 1 3 = π 3 . 1
4. Find the volume obtained by rotating the region between the x -axis, the y -axis, and the line x + y = 1 around the line x = - 2. Answer: I plan to use the shell method. Since the curve is y = 1 - x , the shell based at x has radius x - ( - 2) = x + 2 and height 1 - x , so the volume is Z 1 0 2 π ( x + 2)(1 - x ) dx = 2 π Z 1 0 ( - x 2 - x + 2 ) dx = 2 π - x 3 3 - x 2 2 + 2 x 1 0 = 2 π - 1 3 - 1 2 + 2 = 2 π 7 6 = 7 π 3 . 5. Calculate the surface area obtained by rotating y = x around the x -axis for 3 / 4 x 15 / 4. Answer: Since dy dx = 1 2 x , the surface area integral is Z 15 / 4 3 / 4 2 π x s 1 + 1 2 x 2 dx = 2 π Z 15 / 4 3 / 4 x r 1 + 1 4 x dx = 2 π Z 15 / 4 3 / 4 r x + 1 4 dx. Let u = x + 1 4 . Then du = dx and the above integral is equal to 2 π Z 4 1 u du = 2 π 2 3 u 3 / 2 4 1 = 2 π 16 3 - 2 3 = 28 π 3 . 6. (a) Find the area between y = 3 sin( πx ) and y = 6 x for x > 0. 0.2 0.4 0.6 0.8 1.0 1 2 3 4 5 6 Answer: First, we need to find the points where the curves intersect. The curves intersect when 3 sin( πx ) = 6 x, meaning that sin( πx ) = 2 x, which happens when x = 0 or x = 1 / 2. Therefore, the area of the given region will be given by the definite integral Z 1 / 2 0 (3 sin( πx ) - 6 x ) dx = - 3 cos( πx ) π - 3 x 2 1 / 2 0 = 0 - 3 4 - - 3 π - 0 = 3 π - 3 4 . 2
(b) What is the volume obtained by rotating this region around the x -axis? Answer: My plan is to use the washer method. The outer radius of the washer at x is 3 sin( πx ), whereas the inner radius is 6 x , so the area of the washer at x is A ( x ) = πR 2 - πr 2 = π (3 sin( πx )) 2 - π (6 x ) 2 = 9 π sin 2 ( πx ) - 36 πx 2 . Therefore, the volume of the solid is Z 1 / 2 0 A ( x ) dx = Z 1 / 2 0 ( 9 π sin 2 ( πx ) - 36 πx 2 ) dx = 9 π Z 1 / 2 0 sin 2 ( πx ) dx - 36 π Z 1 / 2 0 x 2 dx = 9 π Z 1 / 2 0 1 - cos(2 πx ) 2 dx - 36 π x 3 3 1 / 2 0 = 9 π x 2 - sin(2 πx ) 4 π 1 / 2 0 - 3 π 2 = 9 π 1 4 - 3 π 2 = 9 π 4 - 3 π 2 = 3 π 4 .

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