Math (Calculus II) Final Eam Form A Fall 22 RED KEY Part I: Multiple Choice Mark the correct answer on the bubble sheet provided.. Which of the following series converge absolutely? ) ( ) n 2) n 2 n ( ) n n n ) n n a) None b) c) 2 d) e), 2 f), g) 2, h), 2, f) 2. Find the radius of convergence of the series n n!( 2) n n 2 n. a) b) c) 2 d) e) 2/ f) 4/ g) / h) 2 i) j) a). What is the coefficient of in the Maclaurin series of e 2? a) b) c) d)! g) 5 5! g) e 2 ( 2 ) k k! k. The coefficient is ( )5 5! e) 25 25! h) 5 5! f) 5 5 i) Diverges ( ) k 2k. When k 5, we obtain the term containing k! k 5 5!.
Fall 22 Math Final Eam Page 2 of 4. When a particle is located a distance feet from the origin, a force of 2 + 2 pounds acts on it. How much work (measured in foot-pounds) is done in moving it from to? a) 6 b) 9 c) 5/ d) 7/ e) 6/ f) 45/2 g) 27/2 h) 5/2 i) c) W ( 2 +2) d ( /+ 2 ) (9+9) (/+) 7 / 5 5/ 5. The region between the curve y and the -ais for < is rotated about the -ais to p form a solid of revolution. For which positive values of p does this solid have finite volume? a) < p b) < p < c) < p d) < p < /2 e) /2 < p f) < p < 2 g) 2 < p h) The volume is infinite for all positive p. d) ( ) 2 d The volume is V π d π, where the integral is improper since p 2p p >. The integral converges when p < /2 but diverges when p /2. 6. Evaluate the integral π/2 sin () cos 2 () d. a) /5 b) 2/5 c) /2 d) / e) /5 f) π g) π/ h) π/2 i) b) Let u cos, du sin d. Then π/2 7. Evaluate the integral sin () cos 2 () d d + 4 2. π/2 ( cos 2 ) cos 2 sin d (u 2 u 4 ) du 5 2 5. ( u 2 )u 2 ( du) 2
Fall 22 Math Final Eam Page of a) b) c) arcsin(/2) d) π e) 2π f) π/2 g) arctan(π) h) f) 2 i) 2 Set u 2 and du 2 d. Then d du/2 and d + 4 2 du/2 + u 2 2 arctan(u) (π/2) ( π/2) π/2. 2 2 8. Which integral represents the length of the curve y sin + cos, π/4? (You might need to set up an integral and do a short calculation.) a) c) e) π/4 π/4 π/4 b) π/4 π/4 2 + 2 cos sin d b) 2 2 cos sin d π/4 2 cos sin d d) 2 cos sin d π/4 cos sin d f) + cos sin d If f() sin + cos, the length of the curve is π/4 π/4 + [f ()] 2 d + [cos sin ]2 d π/4 2 2 cos sin d 9. Which integral represents the area of the surface obtained by rotating the curve about the y-ais. y e, y 8 + cos 2 2 cos sin + sin 2 d a) c) e) 8 ln 8 8 c) 2π + e 2 d b) 2π + e 2 d d) 2πe + e d f) ln 8 ln 8 8 2π + e 2 d 2πe + e 2 d 2πe + e 2 d
Fall 22 Math Final Eam Page 4 of For the curve y e, y 8 the values of satisfy ln 8. Since rotation is about the y-ais, the radius is. ln 8 2πds 2π ln 8 + (dy/d) 2 d 2π + e 2 d. If (, ȳ) is the centroid of the region bounded by the line y and the parabola y 2, what is ȳ? a) b) /2 c) / d) 2/ e) /4 f) /4 g) /5 h) 2/5 i) /5 h) The area is A (/2)/(/6)[ ] (2/5) 2/5 5 ( 2 ) d 2 6. Then ȳ A 2 [()2 ( 2 ) 2 ]d. A curve is parametrized by the equations 6 sin t and y t 2 + t. Find the slope of the line that is tangent to this curve at the point (, ). a) b) c) /2 d) 2 e) / f) g) /6 h) 6 i) Undefined g) 2. Determine the eact value of the geometric alternating series: 7 7 2 + 7 7 4 + a) /2 b) /4 c) 7/6 d) 7/4 e) /8 f) 7/8 e) /7 ( /7) /7 8/7 /8 4
Fall 22 Math Final Eam Page 5 of. Which of the following three tests will establish that the series n n(n + 2) converges? ) Comparison Test with n 2) Limit Comparison Test with ) Comparison Test with n 2n 2 n n 2 n 2 a) None b) c) 2 d) e), 2 f), g) 2, h), 2, g) Part II: Written Response Neatly write the solution to each problem. Complete eplanations are required for full credit. 4. (6 points) Find the volume of the solid obtained by rotating the region bounded by y 2 and y about the vertical line 2. We will compute volume by using cylindrical shells as in the figure below: The cylindrical shell of radius 2 has height 2. So, V 2π 2π radius height d 2π ( 2 + 2) d ( ) 4 2π 4 + 2 π 2 (2 )( 2 ) d ( ) ( ) 4 2π 4 + 2 2π 4 + 2 5
Fall 22 Math Final Eam Page 6 of 5. (6 points) Evaluate sin() d. Use integration-by-parts: u dv uv v du. sin() d }{{} u ( cos() ) }{{} v cos() + sin() 9 ( cos() ) }{{} v + C d }{{} du where { u du d dv sin() d v cos() 6. (6 points) Evaluate 42 d. Use the substitution sin θ and d cos θ for π/2 < θ < π/2 with the 2 2 associated trigonometric diagram: 2 θ 4 2 42 d sin 2 θ cos θ dθ cos θ cos θ dθ 2 2 cos 2 θ dθ ( + cos(2θ)) dθ 2 2 2 4( θ + 2 sin(2θ)) + C 4( θ + sin θ cos θ) + C 4( arcsin(2) + 2 4 2 ) + C 4 arcsin(2) + 2 4 2 + C d 7. (6 points) Evaluate the integral 2 +. First, find the partial fraction decomposition of the integrand: 2 + ( + ) A + B + A( + ) + B Evaluating at gives A. Evaluating at gives B. So ( d 2 + ) d ln ln + + C ln + + + C 6
Fall 22 Math Final Eam Page 7 of The definite improper integral is d B 2 + lim d B 2 + lim ln B + ( ( ) ( )) B lim ln ln B B + + ( ) ln() ln + ln(2) 2 8. (6 points) Let s(n) n k why this choice of n is large enough. ln(2) k. Find a large enough value of n such that s(n) 2, and justify Hint: Think about the geometric reasoning used in the proof of the Integral Test. Since k is a decreasing positive function of positive k, we regard the term / k as a rectangle of height / k and width between k and k +. Then we have s(n) n k n+ k d 2 n+ B 2 n + 2. We ll choose n large enough such that the lower bound for s(n) is 2. 2 n + 2 2 n + n + n + 2 n 2 If we choose n 2, then s(n) 2. [By doing a computer calculation, we find that s(4) 9.946 and s(5) 2.8. Thus, the smallest correct value of n would be n 5. However, this would be difficult to check by hand. The upper bound in the integral above could be replaced with a smaller value like n resulting in a slightly cruder, but correct, lower bound.] 9. (6 points) Determine the interval of convergence for the power series The series clearly converges if /2. (2 )n Assume /2. If a n 5 n n, then lim a n+ n a n lim (2 ) n+ n 5 n+ n + 5 n n (2 ) n lim n n 2 n 5 n + (2 ) n 5 n n. 2 5 7
Fall 22 Math Final Eam Page 8 of Then 2 < 5 < 2 < 5 2 < <. 5 By the Ratio Test the series converges for ( 2, ) and diverges for < 2 or >. If 2, is a convergent alternating series. If, n n (2 ) n 5 n n (2 ) n 5 n n diverges by the p-series test with p /2. The interval of convergence is [ 2, ) n n ( ) n n 2. (6 points) Assuming < <, evaluate the definite integral the answer using summation notation. n du as a power series. Epress + u7 In the integrand, u <. Hence we may use the formula for geometric series r r n where r u 7. n du + u 7 n ( ) ( ) n u 7n du n ( u ( ) n 7n+ ) 7n + n ( ) n 7n+ 7n + ( ) n u 7n du n n ( ) ( ) n 7n+ 7n + 7n+ 7n + 2. (7 points) Find the Taylor series for the function f() centered at the value a. Epress the answer using summation notation. [First Solution] Use the theorem on Binomial Series: In this case, for <, we have ( ) /2 + ( ) ( ) n n ( ) r r(r )(r 2) (r n + ) where, for real r,. n n! ( ) r [The tetbook does use the notation, which is consistent with the traditional notational convention that an empty product equals.] n 8
Fall 22 Math Final Eam Page 9 of [Second Solution] f() /2 f () 2 /2 f () ( 2 2 )/2 2 f (n) () ( )( 2) ( n + 2 2 2 2 )/2 n Then f() and for n f (n) () ( )( 2) ( n + ) 2 2 2 2 The Taylor series centered at a is f (n) ()( ) n n n! + n ( )( 2) ( n + ) 2 2 2 2 ( ) n n! In a (mostly futile) attempt at simplification, we could manipulate the numerator in the previous formula as follows: ( )( 2) ( n + ) ( )( )( 2n ) ( ) 2 2 2 2 2 2 2 2 ( )n 5 (2n ) 2 n ( )n (2n )! 2 n 2 4 6 (2n 4) ( )n (2n )! 2 2n 2 (n 2)! So, in the interval <, we also could write 2 + n2 ( ) n (2n )! ( 2 2n 2 )n (n 2)!n! 22. (6 points) Find the area of the region that lies inside the first curve and outside the second curve: r cos θ, r + cos θ. 9
Fall 22 Math Final Eam Page of 2. (6 points) A trough is full of water. Its end is shaped like the shaded region in the picture. The boundaries of the region are the curves y and y + /2 for. If the pressure at depth d is P δd, where δ is a constant and d is measured in meters, set up a definite integral for the hydrostatic force F against the end of the trough. [Note: Set up an integral for F, but don t evaluate the integral. The answer will involve δ.] y y Solving for in terms of y for the left and right boundary curves gives: y + ( ) /2 ( + y) 2 y + /2 ( + y) 2 A thin horizontal strip at position y, where y, is at depth d y y. The width of this strip is ( + y) 2 [ ( + y) 2 ] 2( + y 2 ). Then F δd dy δ( y)2( + y) 2 dy 2δ y( + y) 2 dy END OF EXAM