Millersville University Name Answer Key Mathematics Department MATH 2, Calculus II, Final Examination May 4, 2, 8:AM-:AM Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of mathematical concepts we have covered. Please show all wor and write out your wor neatly. Answers without supporting wor will receive no credit. The point values of the problems are listed in parentheses.. 8 points) Find the area bounded between the curves. The curves intersect when with x 3 2x 2 x for x.. y y x 3 and y 2x 2 x. 2x 2 x x 3 x 3 2x 2 + x xx 2 2x + ) xx ) 2 x and x.8.6.4.2.2.4.6.8. x A x 3 2x 2 x)) dx x 3 2x 2 + x) dx 4 x4 2 3 x3 + 2 x2 ) 4 2 3 + 2 2
2. 6 points each) Evaluate the following integrals. a) 3x sin2x) dx Integrate by parts letting u 3x du 3 dx v 2 cos2x) dv sin2x) dx. Applying the integration by parts formula we obtain 3x sin2x) dx 3 3 2 x cos2x) + cos2x) dx 2 3 2 x cos2x) + 3 4 sin2x) + C b) cos 4 x sin 3 xdx Using the trigonometric identity sin 2 x cos 2 x we can write cos 4 x sin 3 xdx cos 4 x sin 2 x sin xdx cos 4 x cos 2 x) sin xdx. Integrate by substitution by letting u cos x du sin xdx yields cos 4 x sin 3 xdx cos 4 x cos 2 x) sin xdx u 4 u 2 ) ) du u 6 u 4 ) du 7 u7 5 u5 + C 7 cos7 x 5 cos5 x + C.
c) xe 3x dx Note that using integration by parts with u x du dx v 3 e 3x dv e 3x dx, we find the indefinite integral xe 3x dx 3 xe 3x + 3 e 3x dx 3 xe 3x 9 e 3x + C. Now we may evaluate the improper integral. R xe 3x dx lim xe 3x dx R lim R 3 xe 3x ) R 9 e 3x lim 3 Re 3R 9 e 3R + 3 e 3 + ) 9 e 3 R lim R R 3e 3R 9e + 4 ) 3R 9e 3 4 9e 3.2228 3. 6 points) Find the polar form of the equation in rectangular coordinates x 2 + y 2 x + y. r cosθ) 2 + r sin θ) 2 r cos θ + r sin θ r 2 cos 2 θ + sin 2 θ) rcosθ + sin θ) r 2 rcosθ + sin θ) r cosθ + sin θ
4. 8 points) A 55-lb force stretches a spring 3 inches beyond its natural length. How much wor is done stretching the spring 6 inches beyond its natural length? If the spring obeys Hooe s Law then F x where x is the distance the spring is deformed by a force F and is called the spring constant. In this case Thus the wor done is 55 3 55 3. 6 55 W 3 xdx 275 6 3 x2 33 inch-lbs. 5. 8 points) Find the radius and interval of convergence of the series Using the Ratio Test we see that + x + 2) + 3 lim + x + 2) 3 3 x + 2). lim lim + + x + 2. 3 Thus the power series converges absolutely when 3 x + 2) + 3 + x + 2) x + 2) 3 x + 2 < x + 2 < 3 5 < x <. 3 The radius of convergence is r 3. At x 5 the series becomes 3 5 + 2) ) 3 3) which diverges by the th term test. At x the series becomes 3 + 2) 3 3) which diverges by the th term test. Thus the interval of convergence is 5 < x <.
6. 4 points) Set up a definite integral which gives the volume of the solid of revolution generated when the region bounded between the curves y x 2, y, and x is revolved around the y-axis. You do not need to evaluate the definite integral...8.6 y.4.2...2.4.6.8. x Using the Method of Shells we may set up the integral V 2π xx 2 ) dx 2π x 3 dx π 2. 7. 6 points each) Determine whether the following series converge absolutely, converge conditionally, or diverge. State the names of all convergence or divergence tests you use. a) 2 + 4 3 + 3 + 2 This is a positive term series. Suppose a the Limit Comparison Test we see that 2 + 4 3 + 3 + 2 and b. Applying Since a lim lim b 2 +4 3 +3+2 3 + 4 lim 3 + 3 + 2. b diverges Harmonic Series) then the original series also diverges.
b) ) 3 2 Applying the Ratio Test we see that lim + 3+ ) +) 2 ) 3 2 so the series diverges. lim 3 + 2 ) 3 lim 3 + ) 2 2 + ) 2 ) 3 > c) ln + ) 3/2 This is a positive term series. If we apply the Integral Test we see that xln x + ) dx u 3/2 du 3/2 lim R R lim R lim R 2 <. u 3/2 du 2 u /2 2 2 ) R R /2 Since the improper integral converges the infinite series converges absolutely as well. )
8. points) Find the area of the region that is inside the circle r 3 and outside the cardioid r 3 3 cos θ. Please include a plot of both curves on the same set of axes. Label the points of intersection of the two curves. 6 y 4 Θ Π 2 2 6 4 2 2 4 6 x 2 4 Θ Π 2 6 The curves intersect at θ ±π/2. The area between the curves is π/2 A [ 3) 2 3 3 cosθ) 2] dθ 2 π/2 π/2 [ 9 9 + 8 cosθ 9 cos 2 θ ] dθ 2 π/2 9 π/2 [ 2 cosθ cos 2 θ ] dθ 2 π/2 9 π/2 [2 cosθ ] 2 π/2 2 + cos2θ)) dθ 2 sin θ θ2 4 ) π/2 sin2θ) 9 2 9 2 9 2 2 sin π 2 π/2 2 4 sinπ) 2 π ) 9 2 + π ) 4 2 4 8 9π 4.934 π/2 ) 9 2 2 sin π 2 + π/2 2 + ) 4 sinπ)
9. points) Find the first four terms of the Taylor series for fx) sin x about c π/4. f ) x) f ) f ) π/4) π/4)! sin x 2 2 cosx 2 2 2 sin x 2 2 2 3 cos x 2 6 2 Thus sin x + x π ) 2 2 4 2 x π ) 2 2 4 6 x π ) 3 +. 2 4
. 6 points) Find the partial fraction decomposition of x 2 3x + 4 x 3 + x x2 3x + 4 xx 2 + ) x 2 3x + 4. x 3 + x Equating numerators we have the equation: A x + Bx + C x 2 + Ax2 + ) + Bx + C)x xx 2 + ) x 2 3x + 4 Ax 2 + ) + Bx + C)x If we let x then we see that A 4. Substituting this in the equation we have x 2 3x + 4 4x 2 + ) + Bx + C)x 3x 2 3x Bx + C)x 3x 3 Bx + C and we see that B 3 and C 3. Thus the partial fraction decomposition is x 2 3x + 4 x 3 + x 4 x 3x + 3 x 2 +.. 4 points) Set up a definite integral for the arc length of the curve generated by the parametric equations below with t. x t 2 cost y t 2 sin t You do not need to evaluate the definite integral. s ) b 2 dx + a ) 2 dy dt dt dt 2t cost t 2 sin t) 2 + 2t sin t + t 2 cost) 2 dt 4t 2 cos 2 t 4t 3 costsin t + t 4 sin 2 t + 4t 2 sin 2 t + 4t 3 costsin t + t 4 cos 2 tdt 4t 2 cos 2 t + sin 2 t) + t 4 cos 2 t + sin 2 t)dt 4t2 + t 4 dt t 4 + t 2 dt