Math 4, Final Exam. /7/0. No notes, calculator, or text. There are 00 points total. Partial credit may be given. Write your full name in the upper right corner of page. Number the pages in the upper right corner. Do problem on page, problem on page, etc. Circle or otherwise clearly identify your final answer.. 6 points: Evaluate the integral: x e x. Name: Solution: Integrate by parts. Let u = x and dv = e x. Then we have du = x and v = e x. It follows that x e x = x e x e x x = x e x + xe x. We integrate by parts a second time. Let u = x and dv = e x. Then we have du = and v = e x. It follows that x e x = x e x + xe x e x = x e x xe x + e x = x e x xe x e x + C.. 6 points: Evaluate the integral: sin 3 θ cos 4 θ dθ. Solution: Make the u-substitution u = cos θ. The we have du = sin θ dθ. It follows that sin 3 θ cos 4 θ dθ = sin θ cos 4 θsin θ dθ = cos θ cos 4 θsin θ dθ = cos 4 θ cos 6 θsin θ dθ = u 4 u 6 du = u 6 u 4 du = u7 7 u5 5 + C = cos7 θ cos5 θ + C. 7 5
3. 6 points: Evaluate the integral: x 5 x. Solution: Make a θ-substitution. Since the integrand contains an expression of the form a x, we substitute x = a sin θ = 5 sin θ. To see this, one can also draw an auxiliary right triangle with hypotenuse 5, base angle θ, side adjacent θ equal to 5 x, and side opposite θ equal to x. We have x = 5 sin θ = = 5 cos θ dθ, x = 5 sin θ. Further, from the auxiliary triangle, we deduce that 5 x = 5 cos θ. Alternatively, one can see this by substituting x = 5 sin θ and using the identity sin θ = cos θ. The substitution yields x 5 = sin θ5 cos θ dθ cos θ = 5 sin θ dθ = 5 dθ 5 x = 5 θ cos θ dθ = 5 = 5 5 θ sin θ cos θ + C = x 5 x = 5 sin x 5 5 cos θ θ sin θ + C = 5 θ sin θ cos θ + C x sin x 5 x 5 5 + C 5 + C. 4. 6 points: Evaluate the integral: x + 3x x + x. Solution: We decompose the integrand into partial fractions: x + 3x x + x = A x + + B x + + C x. Clearing denominators multiplying by x + x gives x + 3x = Ax + x + Bx + Cx + Substituting x = gives 4 = 4C; hence, we have C =. Similarly, substituting x = gives = B; it follows that B =. Substituting x = 0 gives 0 = A B + C = A; therefore, we see that A = 0. Alternatively, we expand x + 3x = Ax + x + Bx + Cx + = Ax + Bx + Cx + x + = A + Cx + B + Cx + A B + C.
Comparing coefficients of x 0, x, and x yields: A + C =, B + C = 3, 3 A B + C = 0. There are many ways to obtain A, B, C. For example, equation 3 gives C = A + B. Substituting in equations and gives A + A + B = A + B =, B + A + B = A + 3B = 3 Subtracting these equations gives B = ; hence, we have B =. Substituting this value in equation gives C = ; it follows that C =. Substituting this value in equation yields A = 0. Returning to the integral, we compute x + 3 x = x + x + + = x + + x = x + + ln x + C. x 5. 6 points: Determine whether the integral If it converges, find its value. Solution: This is an improper integral. We compute e t xln x t xln x t e ln t = =. = lim t e is convergent or divergent. xln x ln t u du ] ln t t u u= 6. 5 points: Let R be the region enclosed by y = x, y = x, x = 0 the y-axis. Set up an integral or sum of integrals giving the volume obtained by revolving around the line x =. Do not evaluate the integral. Make a sketch to help justify your reasoning. Solution: The shell method is best suited for this computation: AreaR = π Alternatively, one can apply the washer method: AreaR = π 0 0 x + x x. [ y + ] dy + π [3 y ] dy. 3
7. 5 points: Let R be the region enclosed by y = /x, y =, y =, x = 0 the y-axis. Set up an integral or sum of integrals giving the volume obtained by revolving around the y-axis. Do not evaluate the integral. Make a sketch to help justify your reasoning. Solution: The method of disks is well-suited for this computation: Alternatively, the shell method gives: AreaR = π AreaR = π / 0 y dy. x + π x / x 8. 5 points: Use a comparison test to determine whether the series converges or diverges. n= n + 3 n + n + 5 Solution: The limit comparison test LCT is well-suited for this problem. Observe that when n is large, we have n + 3 n + n + 5 Hence, we compare with. We have n3/ a n lim n b n n n n = n/ n = n 3/. n + 3 n n 3/ n + n + 5 n3/ = > 0. n n Now, since is convergent p-series with p = 3/, we conclude by the LCT that n3/ an converges. Alternatively, one can use the comparison test CT: n + 3 n + 3n n + n + 5 < = n/ = n n n 3/ = n + 3 n + n + 5 < n 3/ = n 3/. Now, since the sum on the right converges, we conclude by the CT that the sum on the left also converges. 4
9. 0 points: Determine whether the series n n!n! converges absolutely, n! n= converges conditionally, or diverges. With a n = n n!n!, compute a n+ /a n n! to get started. Solution: Apply the ratio test. We have L a n+ n a n n +! n +! n [n + ]! n + n + n 4n + 6n + = 4 <. Hence, the series converges absolutely. n! n!n! n + n n + n + 0. 0 points: Find the radius of convergence and the interval of convergence of the power x 5 n series. Test the endpoints of the interval, if necessary. n + Solution: Apply the ratio test: L x 5 n+ n n + + n + x 5 n n + = x 5 lim n n + 3 < x 5 < 4 < x < 6 < x < 3. = x 5 < Hence, the radius of convergence is / it is half of the length of the interval of convergence. It remains to test endpoints. x =. The series is n n. Note that the sequence {/n + } is decreasing n + and that lim = 0. The alternating series test AST implies that the series n + converges. x = 3. The series is. We observe that n + n + > n + n = 4n = n + > 4n = 4 n. The series on the right diverges harmonic; hence, the comparison test CT implies that the series on the left diverges. Alternatively, one could apply the limit comparison test also a comparison with the harmonic series. We conclude that the interval of convergence is [, 3. 5
. 5 points: Let fx = e x. a 5 points: Write down the Taylor series for fx at a = 0. This is sometimes called the MacLaurin series for fx. You do not have to give justification. You can simply write it down from memory, or you can derive it from scratch. Solution: e x = x n n! = + x + x + x3 6 + x4 4 +. e x x x b 5 points: Use your answer from a to compute the limit lim. You are not allowed to use L Hôpital s rule. Solution: e x x x lim x3 6 x 0 4 + x 5! + x 4 + x5 + 4 5! = 4. x3 6 x 4 + x + 4 5! ln n c 5 points: Use part a to evaluate the sum. n! Give an exact numerical value. Solution: e ln = e ln =.. 0 points: Let fx = /x. a 6 points: Compute the Taylor polynomial of degree 3 for fx centered at the point a =. Solution: We have: fx = /x = x, f x = x 3, f x = 6x 4, f 3 x = 4x 5. In turn, we have Hence, we find that f =, f =, f = 6, f 3 = 4. T 3 x = + x + 6 x + 4 x 3 6 = x + 3x 4x 3. 6
b 7 points: i. Find the Taylor series for fx centered at the point a =. For this purpose, you will need to compute a formula for the nth coefficient. Solution: From part a, we see that the nth coefficient is n n +. Hence, the Taylor series is n n + x n. ii. What is the radius of convergence? Solution: We use the ratio test: L n + x n+ n nx n = x lim + = x. n n Hence, the series converges absolutely for x <. It follows that the radius of convergence is. c 7 points: Let gx = /x 3. Note that gx = f x. i. Use your formula from part b to give the Taylor series for gx centered at a =. How does one differentiate a power series? Solution: We compute f x = d n n + x n = n d n + x n = n nn + x n n= = n+ n + n + x n n n + n + = x n n + = n x n ii. What is the radius of convergence? Solution: The radius of convergence is invariant under differentiation. Hence, it is equal to one. 7