Name: Answer Key David Arnold. Math 50B Integral Calculus May 13, Final Exam

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1 Math 5B Integral Calculus May 3, 7 Final Exam Name: Answer Key David Arnold Instructions. (9 points) Follow the directions exactly! Whatever you are asked to do, you must do to receive full credit for the problem. The solution to each problem must be placed on a separate sheet of paper. Place the number of the problem on your solution paper where it will be visible when you staple your solutions together. When you complete the exam, arrange your solutions in order, then staple this exam on the top of your solutions. Good luck! ( pts ). Find the volume generated when the region bounded by the curves y x and y x in the first quadrant is rotated about the y-axis. Include a detailed sketch with your solution. Solution: Start by drawing an image. We take a rectangle perpendicular to the axis of rotation. Rotating about the y-axis forms a disk washer, with inner radius y and outer radius y. Hence, the volume of the washer (with thickness or height dy) is: dv π ( y) dy πy dy π ( y y ) dy Because y was chosen between and, we now integrate over that region. V π(y y ) dy [ π y 3 y3 [ π ] 3 π 6 ]

2 Math 5B Integral Calculus/Final Exam Page of 9 Name: Answer Key ( pts ). Determine the area of the region bounded by the curve r cos θ. Include a detailed sketch with your solution. Solution: First, sketch the graph of the cardioid r cos θ. y x Then we sketch one sector with radius r cos θ and angle dθ, which has area equal to: da (4 + 4 cos θ) dθ [4( + cos θ)] [ 6( + cos θ) ] 8( + cos θ + cos θ) dθ [ ] + cos θ 8 + cos θ + dθ 4 [ + 4 cos θ + + cos θ] dθ 4 [3 + 4 cos θ + cos θ] dθ Because of the symmetry of the cardioid, we can find the total area by summing the sectors from θ to θ π and doubling the result. π A 4 [3 + 4 cos θ + cos θ] dθ [ 8 3θ + 4 sin θ + ] π sin θ 8 [(3π + + ) ( + + )] 4π ( pts ) 3. Determine the length of the curve defined by x e t cos t and y e t sin t on the time interval t π. Include a detailed sketch with your solution. Solution: First, a sketch using Mathematica. ParametricPlot[{Exp[t] Cos[t], Exp[t] Sin[t]}, {t,, \[Pi]}, AxesLabel -> {"x", "y"}]

3 Math 5B Integral Calculus/Final Exam Page 3 of 9 Name: Answer Key y x If x e t cos t, then: Secondly, if y e t sin t, then: et cos t e t sin t e t (cos t sin t) dy et sin t + e t cos t e t (sin t + cos t) Next: ( ) + ( ) dy [ e t (cos t sin t) ] [ + e t (cos t + sin t) ] e t (cos t cos t sin t + sin t) + e t (cos t + cos t sin t + sin t) e t ( + ) e t Now we can find the length of the path. But e t is always positive, so: L π π ( ) + e t π π e t e t π e t [e π ] ( ) dy e x sin x. An answer without the appro- ( pts ) 4. Use integration by parts to evaluate the indefinite integral priate steps shown will be unacceptable. Solution: First, set up an integration by parts table. D I + sin x e x cos x e x + sin x e x

4 Math 5B Integral Calculus/Final Exam Page 4 of 9 Name: Answer Key Thus: e x sin x e x sin x e x cos x e x sin x e x sin x e x cos x e x sin x ex (sin x cos x) + C e x sin x ( pts ) 5. Evaluate the definite integral unacceptable. Solution: Let: Then: 3/ 3/. An answer without the appropriate steps shown will be 9 x x 3 sin θ 3 cos θ dθ 9 x 3 cos θ dθ 9 9 sin θ 3 cos θ dθ 3 sin θ cos θ dθ cos θ cos θ dθ cos θ However, because we let x 3 sin θ and the range of the inverse sine is in the first and fourth quadrants where the cosine is always positive, we can remove the absolute value bars and continue. cos θ dθ cos θ dθ Since x 3 sin θ, this means that θ sin θ x 3 or θ sin x 3 Thus: 3/ 3/ x 3/ 9 x sin 3 3/ sin ( sin ) π ( 6 π ) 6 π 3

5 Math 5B Integral Calculus/Final Exam Page 5 of 9 Name: Answer Key x ( pts ) 6. Evaluate the definite integral x. An answer without the appropriate steps shown will + 3x + be unacceptable. Solution: We first need to do a partial fraction decomposition. x (x + )(x + ) A x + + B x + x A(x + ) + B(x + ) Now, if x, we get B 3. If x, we get A. Thus, we can write: x x + 3x + [ x ] x + [ ln x ln x + ] ( ln + 3 ln 3) ( ln + 3 ln ) 3 ln 3 5 ln ln 33 5 ln 7 3 ( pts ) 7. Consider the recursively defined sequence whose first term is a and the remaining terms are defined by the recursion formula a n+ + a n. (a) Use mathematical induction to prove that a n for all n,, 3,.... Solution: First, a, so the statement a n is true for n. Next, assume that the statement is true for n k. Now we need to show it is true for n k +. a k + a k + + a k 4 + ak 4 a k+ Hence, a n for all n,, 3,.... (b) Use mathematical induction to prove that the sequence is increasing for all n,, 3,.... Solution: First, a + a, so a n a n+ is true for n. Next, assume that the statement is true for n k. a k a k+ Now we need to show that it is also true for n k +. + a k + a k+ + ak + a k+ a k+ a k+ Hence, a n a n+ for all n,, 3,.... That is, the sequence is increasing.

6 Math 5B Integral Calculus/Final Exam Page 6 of 9 Name: Answer Key (c) Explain why the sequence must have a limit, then determine the limit of the sequence. Solution: Because the sequence is increasing and bounded above, it converges. Let lim a n L, then, because {a n+ } is a subsequence of {a n }, it also converges and Thus, we can write: lim a n+ L. lim a n+ lim + an L + L L + L L L (L )(L + ) L, Because this is a sequence of positive terms that are increasing, we choose L as the limit. ( pts ) 8. Consider the power series defined by: n ( 3) n x n n + (a) Use the ratio test to determine the radius of convergence. State your answer using the form The radius of convergence is R, where R is the radius of convergence. Solution: We use the ratio test to write as n. Now, if a n+ a n ( 3) n+ x n+ n + n + ( 3) n x n n + 3 n + x + n 3 + x 3 x n 3 x < x < 3 or equivalently, if 3 < x < 3, then the series converges. Thus, we have the radius of convergence is R /3. (b) Use a test of your own choice to determine whether the series converges or diverges at the end points of the interval of convergence, then use interval notation to state the interval of convergence.

7 Math 5B Integral Calculus/Final Exam Page 7 of 9 Name: Answer Key Solution: Now we test the endpoints of the interval of convergence. First, if x /3, then the series becomes: ( 3) n x n ( 3) ( ) n n 3 n + n + n n n (n + ) / We will compare the following series. n (n + ) / and n n / Both have positive terms a n (n + ) and b / n. Further, note that: n/ The series n a n lim lim b n lim lim lim constant, by the limit comparison test, the series n finite term does not affect the convergence, so n (n + ) / n / n / (n + ) / ( n n + ( + n ) / ) / n is a p-series with p <, so it is divergent. Because lim a n /, a non-zero b n is also divergent. Adding another (n + ) / also diverges. (n + ) / Next, if x /3, the series becomes n ( 3) n x n n + ( 3) ( n 3 n + n n ) n ( ) n (n + ) /. which is an alternating series. Set b n /(n + ) /. Note that:. b n > for all n.. Start with n + > n and apply the square root function to both sides of the inequality. n + > n

8 Math 5B Integral Calculus/Final Exam Page 8 of 9 Name: Answer Key We do not reverse the inequality sign because the square root is an increasing function. Next, we apply the function /x to both sides of the inequality. < n + n We reverse the inequality because the function /x is a decreasing function. Because b n /(n + ) /, we ve shown that b n+ < b n for all n. Therefore, the sequence {b n } is decreasing. 3. Finally, note that lim n + Thus, by the alternating series test, the series converges. Hence, using interval notation, the interval of convergence is: ( 3, ] 3 ( pts ) 9. The goal of this problem is to determine a Maclaurin s series generated by the function f(x) cosh x. Recall the following definitions: cosh x ex + e x and sinh x ex e x (a) First, determine formulae for the nth derivative of f(x), then use your result to evaluate f (n) () for each value of n. Solution: Note that: so: Hence, and D x sinh x cosh x and D x cosh x sinh x, f(x) cosh x f (x) sinh x f (x) cosh x f (x) sinh x.. f (n) (x) cosh x and f (n+) (x) sinh x f (n) () and f (n+) () (b) Use your results from part (a) to write the first eight terms of the series in the form Solution: Thus: f() + f ()! x + f ()! x + f (7) () x 7 + 7! f() + f ()! x + f ()! x + f (7) () x 7 + 7! + x +! x + x 3 + 4! x4 + x 5 + 6! x6 + x 7 +

9 Math 5B Integral Calculus/Final Exam Page 9 of 9 Name: Answer Key (c) Finally, write your Maclaurin s series using formal summation notation. Solution: Thus: cosh x n x n (n)!