Final Exam Review Quesitons

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1 Final Exam Review Quesitons. Compute the following integrals. (a) x x 4 (x ) (x + 4) dx. The appropriate partial fraction form is which simplifies to x x 4 (x ) (x + 4) = A x + B (x ) + C x Dx x + 4, x x 4 = A(x )(x + 4) + B(x + 4) +C(x ) + Dx(x ). Plugging in the following allows you to solve for A, B, C, and D. x = : = 8B, B = x = : 4 = 8A + 4B + 4C, = 8A + 4C, C = A x = : 4 = 5A + 5B +C + D, 9 = A + D, D = A 9 x = : 4 = 5A + 5B + 9C 9D, 8 = 5A + 9(A ) 9(A 9), A =, C =, D =. (b) Then sin 4 (x)cos (x)dx. x x 4 (x ) (x + 4) dx = x dx + (x ) dx + x + 4 dx = ln( x ) x + arctan(x/) +C. sin 4 (x)cos (x)dx = = sin 4 (x)( sin (x))cos(x)dx u 4 ( u )du u-substitution [ u = sin(x) du = cos(x)dx ] = 5 u5 7 u7 +C = 5 sin5 (x) 7 sin7 (x) +C.

2 (c) x 4 (ln(x)) dx. (d) x x + dx. x x + dx = x 4 (ln(x)) dx = 5 x5 (ln(x)) ln(x) x5 dx 5 [ x u = (ln(x)) v = Integrate by Parts 5 x5 du = ln(x) x dx dv = x 4 dx = 5 x5 (ln(x)) + ( ) 5 x4 ln(x)dx [ u = ln(x) v = Integrate by Parts 5 x 5 = = = 5 x5 (ln(x)) 5 x5 ln(x) du = x dx ( 5 x5 = 5 x5 (ln(x)) 5 x5 ln(x) + 5 x5 +C. tan(u) tan (u) + sec (u)du [ x = tan(u) trig-substitution dx = sec (u)du sec(u) tan(u) du = tan (u) sec(u)tan(u)du [ ] v = sec(u) u-substitution dv = sec(u) tan(u)du v dv = v dv v + dv ] dv = 5 x4 dx ) x dx = ln( v ) ln( v + ) +C = ) ( ln + x ( ) ( ) ln + x + +C = + x ln +C + x + ( = ln ( ) + x ) ( ) +C = ln + x ln( x ) +C x ] ]

3 . Determine the values of p for which the following integrals converge. (a) x p dx { converges p < converges p. (b) x p dx { converges p > converges p.. Determine if each of the following integrals is convergent or divergent. (a) x(x + ) dx. You should break the integral in two parts, and write them as limits using the definition of improper integrals, dx = lim x(x + ) t + r dx + lim x(x + ) r x(x + ) dx The first limit exist since the function behaves like x near, and the second limit exists because the function behaves like x/ as x. Then the integral converges. (b) ( + x ) dx. You should again break this into two integrals to handle the and, ( + x dx = lim ) t t r ( + x dx + lim ) r ( + x ) dx. Since the function behaves like x as x ±, it follows that the limits above exist and that the integral converges. (c) (x ) dx. You should break this integral into three pieces, t dx = lim (x ) t dx + lim (x ) r + r s dx + lim (x ) s (x ) dx. The last limit exists, but the first two are infinite. Therefore the integral is divergent.

4 4. Let f (x) = sin(x) and g(x) = sin(x). Consider the region R bounded by f (x) and g(x) on the interval [, π/6]. (a) Sketch the region R. (b) Find the area of the region R. A = π/6 (sin(x) sin(x))dx = ( ) π/6 cos(x) cos(x) = 4 (c) Find the volume of the solid generated by revolving the region R about the x-axis. π/6 V = π (sin (x) sin (x))dx = π = π ( sin(x) ) π/6 4 sin(4x) = π/6 π 6 (cos(x) cos(4x))dx (d) Find the volume of the solid generated by revolving the region R about the y-axis. π/6 V = πx(sin(x) sin(x))dx [ ] u = πx v = cos(x) Integrate by Parts cos(x) du = πdx dv = (sin(x) sin(x))dx = πx(cos(x) π/6 π/6 cos(x)) π (cos(x) ) cos(x) dx π = π π/6 (cos(x) 6 π ) cos(x) dx π = (sin(x) π 6 π 4 ) π/6 sin(x) π = π π π (e) Set up integrals to compute the length of the boundary of the region R. L = π/6 + cos (x)dx + π/6 + 4cos (x)dx + sin(π/) sin(π/6).

5 5. Find all sixth roots of + i. This problem is to find all solutions to z 6 = + i. To do this write both z and + i in exponential form: z = re iθ and + i = e i(π/4+πk) for any integer k. Then the equation becomes r 6 e i6θ = e i(π/4+πk). Now equate the moduli and arguments of this equation to get r 6 = r = 6θ = π 4 + πk θ = π 4 + πk This provides solutions for each integer k. Since there are only 6 solutions to the equation, it is sufficient to plug in k =,,,,4,5 to conclude that the following are all sixth roots of + i z = e i π 4, e i( π 4 + π ), e i( π 4 + π ), e i( π 4 +π), e i( π 4 + 4π ), e i( π 4 + 5π ) 6. Find all solutions to the equation z 5 = i. Use the same technique as above to write z = re iθ and = e i(π/+πk) for any integer k. Then we get r 5 e i5θ = e i(π/+πk), which we solve by r 5 = r = So the five solutions to this equation are 5θ = π + πk θ = π + πk 5. z = e i π,e i( π + π 5 ),e i( π + 4π 5 ),e i( π + 6π 5 ),e i( π + 8π 5 )

6 7. Determine if the following series are absolutely convergent, conditionally convergent, or divergent. (a) (b) n 4 + n n + n= n + n + 4 n 4 + n n + Diverges by the test of divergence since lim n n =. + n + 4 n= n(ln(n)) / Diverges by the integral test since x(ln(x)) ln(t) dx = lim t ln() u dx = lim t (ln(t)) (ln()) =. (c) (d) n= n (.) n Converges absolutely by the ratio test: n= ( ) n (ln(n)) (n + ) (.) n (n + ) L = lim n (.) n+ n = lim n.n =. <. Conditionally convergent by the alternating series test since n and the comparison test since n (ln(n)) for n. (ln(n)) decreases to zero as (e) ( ) n n n= (n + ) Diverges by the test for divergence since lim n ( ) n n (n + ) DNE.

7 8. Suppose a power series a n x n converges when x = and diverges when x = 6. (a) Does the power series converge at x =? Yes (b) Does the power series converge at x = 7? No (c) Does the power series converge at x = 6? Not enough information. (d) What can you say about the radius of convergence of the power series? R Find the Taylor series for the following functions centered at the given point. Also give the radius of convergence. (a) f (x) = xe x ; a =. (b) f (x) = xe x ; a = 4. xe x = x n= x n n! = x n+ n= n! ; R = xe x = (x 4)e x + 4e x = e 4 (x 4)e x 4 + 4e 4 e x 4 = e 4 (x 4) = e 4 n= = 4e 4 + n! (x 4)n+ + n= (c) f (x) = cos(x ); a =. n= ( e 4 (n )! + 4e4 n! 4e 4 n! (x 4)n = n= e 4 ) (x 4) n ; R = n= (n )! (x 4)n + 4e 4 + (x 4) n + 4e 4 (x 4) n! n n! n= n= 4e 4 (x 4)n n! cos(x ) = ( ) n n= (n)! (x ) n ( ) = n n= (n)! x4n ; R =

8 (d) f (x) = x ; a =. x = + x = (e) f (x) = x ; a =. ( x ) = ( x ) n = n= ( ) n n= n+ (x )n ; R = Differentiate the previous series to obtain x = n= ( ) n n+ n(x )n ; R = Then x = ( ) n+ n= n+ n(x ) n ; R =

9 . Consider the curve given by the polar equation r = θ θ. (a) Sketch the curve for θ π. (b) Find the derivative dy dx as a function of θ. Recall that x = r cos(θ), and since for this curve r = θ θ, it follows that x = (θ θ)cos(θ). Similarly y = (θ θ)sin(θ). Then and x = (θ θ)cos(θ) y = (θ θ)sin(θ) dy dy dx = dθ dx dθ dx dθ = (θ )cos(θ) (θ θ)sin(θ) dy dθ = (θ )sin(θ) + (θ θ)cos(θ) = (θ )sin(θ) + (θ θ)cos(θ) (θ )cos(θ) (θ θ)sin(θ). (c) Find the equation of the line tangent to the curve at the point (r,θ) = (4π π,π) To find the tangent line, we need a point (x,y ) on the line and the slope given by m = dy. θ=π dx The point (x,y ): The point can be found by the following x = (4π π)cos(π) = 4π π, y = (4π π)sin(π) =. The slope m: The slope can be found by the following m = dy dx = (4π )sin(π) + (4π π)cos(π) θ=π (4π )cos(π) (4π π)sin(π) = 4π π 4π Then using the point slope form for a line, it follows that the equation for the tangent line is y = 4π π 4π (x (4π π)).

10 { k x. A random variable X has distribution f (x) = x x < (a) Find the value of k so that f is a distribution. Then k =. = f (x)dx = ( k dx = lim k ) t = lim k k x t x t t = k. (b) Find the mean. µ = t x f (x)dx = lim t x dx = lim ln(t) =. t The mean for this probability distribution does not exist since this integral does not converge. This type of problem will not be on the final. (c) Find the probability P(4 X ). P(4 X ) = 4 f (x)dx = 4 x dx = 4 =

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