Math 62: Calculus IIA Final Exam ANSWERS December 9, 26 Part A. (5 points) Evaluate the integral x 4 x 2 dx Substitute x 2 cos θ: x 8 cos dx θ ( 2 sin θ) dθ 4 x 2 2 sin θ 8 cos θ dθ 8 cos 2 θ cos θ dθ 8 ( sin 2 θ) cos θ dθ 8 ( u 2 ) du (using the substitution u sin θ) ) 8 (u u + C ( ) 8 sin θ sin θ + C ( 4 x 2 8 ( ) 4 x 2 ) + C 2 24 since sin θ 4 x 2 /2. 2. (2 points) (a) Compute the volume of a region bounded by the curves y x +, y and x and rotated around the y-axis.
Using the shell method we have shells of radius x, thickness dx and height (x + ) x. Therefore V 2πx x dx 2π x5 5 2π 5 (b) Set up the integral for the volume of the region bounded by y x 4, y and x 2 and rotated around line x 2. Use the shell method. Do not evaluate the integral. Using the shell method we have shells of radius (2 x), thickness dx and height x 4. Thus the volume is V 2 2π(2 x)x 4 dx.. ( points) Evaluate the integral arcsin x dx. Integrating by parts with u arcsin x and dv dx, we get du dx x 2 and v x, so that the integral becomes arcsin x dx x arcsin x x dx x 2 Now make the substitution w x 2, so dw 2x dx and x dx dw/2. This means our new integral is x dx x 2 dw 2 w 2 u /2 dw 2 2u/2 C x 2 C, Page 2 of 9
and the original integral is arcsin x dx x arcsin x + x 2 + C. 4. (2 points) (a) Find the partial fraction decomposition of x 2 + x x 2. The fraction is improper so first use long division to write: x + x x 2 + x + x 2. Since the denominator is a difference of squares x 2 (x )(x+) we next seek constants A, B such that: x + x 2 A x + B x + which is equivalent to solving the linear system: A + B A B Adding these equations gives 2A 4 so A 2 and therefore B. Thus: x + x x 2 + 2 x + x +. (b) Write out the form of the partial fraction decomposition of the function x 2 (x + ) (x 2 + ) 2 (x ) Do not determine the numerical values of the coefficients. All the factors are linear except x 2 +, which has discriminant b 2 4ac 4 < (has complex roots ±i) so does not factor over the real numbers. Thus there is a linear factor Page of 9
of multiplicity, an irreducible quadratic factor of multiplicity 2 and a linear factor of multiplicity. So the partial fraction decomposition will look like: x 2 (x + ) (x 2 + ) 2 (x ) A x + + A 2 (x + ) + A 2 (x + ) + B x + C + B 2x + C 2 x 2 + (x 2 + ) + D 2 x. (c) Let f(x) x + 2x + x 2 +. Evaluate f(x)dx. Split the integral: f(x)dx x dx + ln x + 2x x 2 + dx + 2x dx + arctan x x 2 + x 2 + dx Substitute u x 2 + and hence du 2xdx to get: f(x)dx ln x + ln x 2 + + arctan(x) + C. 5. (5 points) The cardioid is the curve defined in polar coordinates by r + cos θ. Find the area of the region bounded above by the cardioid and below by the x-axis. Page 4 of 9
Solution: It is easily verified that the region R bounded above by the cardioid and below by the x-axis is given by R {(r, θ) : r + cos θ, θ π}. We use the formula for area inside a polar curve to compute that the area A of the region R is given by A 2 2 π 4. π π π ( + cos θ) 2 dθ ( + 2 cos θ + cos 2 θ) dθ 2 ( 2 + 2 cos θ + ) cos 2θ dθ 2 2 [ θ 2 + 2 sin θ + sin 2θ 4 ] π 6. (2 points) Find the arc length of the parametric curve x(t) e t cos t, y(t) e t sin t connecting the point (, ) to the point (e 2π, ). Solution: First observe that the points (, ) and (e 2π, ) correspond to t and t 2π, respectively. It follows that the arc length of this curve is given by L 2π 2π 2π 2 Part B 2π (x (t)) 2 + (y (t)) 2 dt (et cos t e t sin t) 2 + (e t sin t + e t cos t) 2 dt e 2t cos 2 t 2e 2t sin t cos t + e 2t sin 2 t + e 2t cos 2 t + 2e 2t sin t cos t + e 2t sin 2 t dt 2e 2t (cos 2 t + sin 2 t) dt 2π 7. (2 points) e t dt 2[e t ] 2π 2(e 2π ). (a) Find a power series representation centered at as well as the radius and interval of convergence for the function f(x) x x + 2. Page 5 of 9
Write f(x) as the sum f(x) x x + 2 x a of a geometric series r n arn [which converges iff r < ] + (x ) This converges if and only if: (x ) + ( ) x r x n ( ) n x (x ) < x < So the radius of convergence is R and the interval of convergence is ( 2, 4). ( ) n (x ) n n (b) Write the following integral as a power series in x. What is the radius of convergence of this power series? x x + 2 dx By the integration theorem: x x 2 dx ( ) n (x ) n dx for x < n n ( ) n (x ) n dx n n ( ) n n with the same radius of convergence R. 8. (2 points) n (n + )(x ) n+ Determine whether the series is absolutely convergent, conditionally convergent, or divergent. n2 ( ) n n ln n n First, consider the series ( ) n n ln n n2 n2 n ln n Page 6 of 9
for absolute convergence. Since n > (ln n) 2 for n 2, n ln n > n. We also know that the harmonic series n2 diverges by the p-series test with p. n Therefore, it follows from the comparison test that the series diverges. Now, we consider the series n2 ( ) n n ln n for conditional convergence. It is an alternating series satisfying lim n n ln n. It is also obvious that n ln n is a decreasing function of n. So by the Alternating Series test, the original series converges. Therefore, the series is a conditionally convergent series. 9. (2 points) Find the radius of convergence and interval of convergence of the series n (x 2) n. n n Solution: We use the ratio test: a n+ a n a n+ a n n+ x 2 n+ n + as n. From the radius of convergence R /. n n + x 2 x 2 x 2 < x 2 <, n n x 2 n Now consider the boundary case x 5/ or x 7/. Plugging x 5/ in original series expression, we get n ( ) n n, Page 7 of 9
which converges by the alternating series test. Plugging x 7/ in original series expression, we get n n, which diverges by the p-series test with p / <. So the interval of convergence is [5/, 7/).. (2 points) Let f(x) x2 + 2x. (a) Find the Taylor series of f(x) centered at x. Write x 2 + 2x x 2 ( 2x) x 2 ( 2x) n n ( 2) n x n+2. n (using geometric series expansion) (b) Find the radius of convergence. One way is to note that f(x) is not defined at x /2. This is a distance of /2 away from the center x. Thus, the radius of convergence will be /2. Alternatively, you can use the Ratio Test: lim ( 2) n+ x n+ ( 2) n x n+2 2x n For the series to converge by the Ratio test, we must have 2x <, which means x < /2. Thus, again, the radius of convergence is /2. (c) Compute f () (). Page 8 of 9
For Taylor series centered at x a, the coefficient of x n is f (n) (a)/n!. Thus, we need the coefficient of x. This happens when n 98. Thus, we have f () ()/! ( 2) 98, and so f () ()! ( 2) 98! 2 98.. (2 points) (a) Determine whether the series n ( ) n+ n 5 is absolutely convergent, conditionally convergent, or divergent. The series converges by the alternating series test. It converges absolutely by the intgeral test or the p-test. (b) Estimate the sum of the series with an accuracy of. /. The alternating series is 2 5 + 5 + 2 + 24 + Its third terms is less that.5 /2, so the sum of the first two terms will give the desired precision. That sum is 2 2.96875. Page 9 of 9