Midterm Exam #1. (y 2, y) (y + 2, y) (1, 1)

Transcription

1 Math 5B Integral Calculus March 7, 7 Midterm Eam # Name: Answer Key David Arnold Instructions. points) This eam is open notes, open book. This includes any supplementary tets or online documents. You are not allowed to work in groups or pairs on the quiz. You are not allowed to enlist the aid of a tutor or friend to help with the quiz. You must answer all of the eercises on your own. You are also not allowed to share your work with your fellow students. Honor Pledge: I promise that the solutions submitted were done entirely by me. I received no help from colleagues, friends, or tutors. I also did not share any parts of my solutions with any of my classmates. Signature: Please staple this cover and honor pledge atop your solutions. Neatness Requirement: Each problem on this eam must be done on a separate sheet of paper. All pencil and paper calculations must be done using pencils. Deductions will be made for any work done in pen. Any mistakes made must be carefully erased with an eraser. If there is any scratching out of work or any disorganized presentation, points will be deducted. pts ). Sketch the region bounded by y and y. Draw the typical element of area you may choose a direction of your choice). Label its endpoints with the coordinates, state the area of this typical element, the set up an integral and integrate to find the eact area of this region. Solution: Start with a sketch. y y, ) y y, y) y +, y), ) Hence, the area of the shaded rectangle is: da y + ) y ] dy Thus, the area bounded by the curves is found by the following integration: A y + ) y ] dy y + y ] y + 8 ) + + ) 9

2 Math 5B Integral Calculus/Midterm Eam # Page of Name: Answer Key pts ). Find the volume of the solid generated when the region between the graphs of the equations f) + and g) over the interval, ] is revolved about the -ais. Use the disk or washer method. Include a detailed sketch with your solution. Solution: Start with a sketch. y y +, / + ) y, ) Now, when we rotate the shaded square about the -ais, we get a washer having an inner radius of and an outer radius of / +. y y + y / + In addition, the thickness is. Thus, the volume of the washer is the volume of the outer disk minus the volume of the inner disk: ) dv π + π ) ] dv π +

3 Math 5B Integral Calculus/Midterm Eam # Page of Name: Answer Key We can now integrate over the interval to find the volume of the figure. ) ] V π + π π ] + + ] + π π + ] 5 69π ] pts ). Find the volume of the solid when the region bounded by y and y is rotated about the line. Use the cylindrical shell method and include a detailed sketch with your solution. Solution: Start with a sketch. Because we are rotating about the line using the cylindrical shell method, we create a rectangle parallel to the ais of rotation. y, ), ) Now, when we rotate about the line, we get a cylindrical shell.

4 Math 5B Integral Calculus/Midterm Eam # Page of Name: Answer Key y, ), ) Therefore, the volume of the cylindrical shell is: Cylindrical Shell Volume Circumference)Height)Width) π ) ) Because we fied a value of between and to start, we continue as follows: Volume π π o π ) ) + ] + ] π + π + ] π ] ] π π 6 ] pts ). Use integration by parts using the tabular integration technique to evaluate the integral: Solution: Use tabular integration by parts. / sin D I + sin

5 Math 5B Integral Calculus/Midterm Eam # Page 5 of Name: Answer Key Hence: Thus: sin sin sin sin + ) / sin + ) / / sin sin + ] / ) sin + π 6 + π + + ) pts ) 5. Evaluate the following definite integral. π cos Solution: We ll use the half-angle identity on this one. π cos Now the half angle identity again π π π cos ) ) + cos + cos + cos ) π π π + cos + ) + cos + cos + + cos ) + cos + cos ) + sin + ] π sin π + sin π + ) ] sin π + + ) π 8

6 Math 5B Integral Calculus/Midterm Eam # Page 6 of Name: Answer Key pts ) 6. Use partial fraction decomposition to evaluate the indefinite integral: + + Solution: We ll start with partial fraction decomposition ) ) + + A + B + + C + A + B) ) + C + ) Setting / give us: 9 + ) C C C C 7 5 Epanding the above epression further gives us: + A + B) ) + C + ) + A A + B B + C + C + A + C) + A + B) + B + C) Hence: A + C A + B B + C Using C 7/5 and the first equation gives us A /5. Using C 7/5 and the third equation gives us B /5. Hence: ln + ) + 5 tan 7 ln + C 5

7 Math 5B Integral Calculus/Midterm Eam # Page 7 of Name: Answer Key pts ) 7. Use trigonometric substitution to evaluate the following definite integral. + Solution: Let tan θ and sec θ dθ. Then: + + tan θ sec θ dθ) sec θ sec θ dθ) sec θ sec θ dθ) Now, because the inverse tangent returns an angle in the first or fourth quadrant, we have the following picture., ) + θ + θ, ) Because the secant is positive in quadrants I and IV, we can remove the absolute value bars. Thus: + sec θ dθ This we will integrate by parts. Hence: D I + sec θ sec θ sec θ tan θ tan θ sec θ dθ sec θ tan θ sec θ dθ sec θ tan θ sec θ dθ sec θ tan θ sec θ tan θ dθ sec θsec θ ) dθ sec θ dθ + sec θ dθ Thus: sec θ dθ sec θ tan θ + ln sec θ + tan θ sec θ dθ sec θ tan θ + ln sec θ + tan θ

8 Math 5B Integral Calculus/Midterm Eam # Page 8 of Name: Answer Key Use the picture above and note that sec θ + and tan θ. Hence: + sec θ tan θ + ln sec θ + tan θ + + ln + + Therefore: + sec θ tan θ + ] ln sec θ + tan θ + ln + ) pts ) 8. Use trigonometric substitution to evaluate the following indefinite integral. 5 Solution: The first step is to complete the square for the denominator ) + + 5) + ) 9) 9 + ) Now, we can write: ) Now, we will use the following u-substitution. u + du Note that u. Making these substitutions, we get: u ) 9 + ) 9 u For the first integral, let s use these substitutions. u 9 u u 9 u w 9 u dw u du ) 9 u

9 Math 5B Integral Calculus/Midterm Eam # Page 9 of Name: Answer Key Therefore, the first integral can be managed as follows: u 9 u For the second integral, let s use these substitutions, u du 9 u dw w w / u sin θ w / dw 9 u ) / ) du cos θ dθ where θ is restricted to π/ θ π/. Substituting these into our second integral, we can write: cos θ dθ) 9 u 9 sin θ) cos θ dθ) 9 9 sin θ cos θ dθ) sin θ cos θ cos θ dθ cos θ cos θ dθ Now, because π/ θ π/, the angle θ lies in quadrant one or quadrant four. In each of these quadrants the cosine is positive, so we may remove the absolute value symbols and continue. cos θ 9 u θ cos θ dθ Because we made the substitution u sin θ, with π/ θ π/, we know we can write: u sin θ dθ sin θ u sin sin θ) sin u ) θ sin u ) The identity sin sin θ) θ is true if and only if π/ θ π/. We can substitute this into our last integration to obtain: θ 9 u sin u ) )

10 Math 5B Integral Calculus/Midterm Eam # Page of Name: Answer Key Now we can substitute equations ) and ) into equation ) to obtain: u 9 + ) 9 u 9 u 9 u ) / sin u + C 9 u sin u + C Recall that our first substitution was u +, so our final answer is: 9 u sin u 9 + ) + C 9 + ) sin + + C 5 sin + + C pts ) 9. Evaluate the following improper integral. Solution: By definition: e ln e ln t lim t e ln Note that if we let u ln and du /), then: ln u du u du Hence: e ln u ln lim t ln lim t lim t ] ] t e ln t ] ln e ] ln t

11 Math 5B Integral Calculus/Midterm Eam # Page of Name: Answer Key pts ). Evaluate the definite integral ) / Solution: The integrand is not continuous over the interval, ]. It is undefined at. So we have an improper integral. The first step is to break it into two pieces. lim ) / b lim b b b + lim ) / a + ) / + lim a + a a ) / ) / lim ) /] b + lim ) /] b a + a lim b ) / ) /] + lim ) / a ) /] b a + lim b lim b b ) / ) /] + lim ] b ) / ) ] lim b ) / + b a + ) / a ) /] + lim ) / a ) /] a + + lim ) / a ) /] a + Now we can evaluate the limits. lim ) / b ] b ) / + + lim ) / a ) /] a ] + + lim ) / ) /] a ] + ) / ) / + + ] + + ) /