Spring 2006 - Nagy's Class - Final Exam

Spring 2006 - Nagy's Class - Final Exam - Print Name: TA...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Print Name: TA Name: Math 20B. Final Exam June 12, 2006 Section Number: Section Time: No calculators or any other devices are allowed on this exam. Write your solutions clearly and legibly; no credit will be given for illegible solutions. Read each question carefully. If any question is not clear, ask for clarification. Answer each question completely, and show all your work. 1. Evaluate the integrals: (a) (8 points) (b) (8 points) x2 cos(x) dx; sin ln(2x) dx. x Show all your work. (a) x2 cos(x) dx = x2 sin(x) - 2x sin(x) dx, 2 - cos(x) dx , cos(x) dx, = x2 sin(x) - -2x cos(x) - = x2 sin(x) + 2x cos(x) - 2 = x2 sin(x) + 2x cos(x) - 2 sin(x) + c, x2 cos(x) dx = x2 sin(x) + 2x cos(x) - 2 sin(x) + c . (b) Substitute u = ln(2x), then du = dx/x, so sin ln(2x) dx = sin(u) du, x = - cos(u) + c, sin ln(2x) dx = - cos ln(2x) + c . x 2. (8 points) Use complex exponentials to compute the integral may leave the result in exponential form. e2x cos2 (x) dx. You e 2x cos (x) dx = = 2 e 2x eix + e-ix 2 2 dx, 1 e2x e2ix + e-2ix + 2 dx, 4 1 e2(1+i)x + e2(1-i)x + 2e2x dx, = 4 e2(1-i)x 1 e2(1+i)x + + e2x + c, = 4 2(1 + i) 2(1 - i) e2(1-i)x 1 e2(1+i)x + + e2x + c . 4 2(1 + i) 2(1 - i) e2x cos2 (x) dx = 3. (10 points) Evaluate the integral x2 2x - 1 dx. - 3x + 2 The roots of x2 - 3x + 2 are x = 2 and x - 1, then x2 - 3x + 2 = (x - 2)(x - 1). 2x - 1 A B (x - 1)A + (x - 2)B = + = x2 - 3x + 2 x-2 x-1 (x - 2)(x - 1) 2x - 1 = (A + B)x - (A + 2B) The solutions are A = 3, B = -1. 2x - 1 3 1 dx = dx - dx, x2 - 3x + 2 x-2 x-1 = 3 ln(|x - 2|) - ln(|x - 1|) + c. x2 2x - 1 dx = 3 ln(|x - 2|) - ln(|x - 1|) + c . - 3x + 2 A + B = 2, A + 2B = 1, 4. (10 points) Use the comparison Theorem to decide whether diverges. If the integral converges, then compute its value. Show your work. 1 dx converges or 1 + x2 x2 1 + x2 1 1 1 + x2 t 1 1 x2 t 1 dx 1 + x2 1 = 1. t 1 dx . x2 dx = lim t x2 1 x-2 dx = lim 1 - 0 dx 1 + x2 1, the integral converges. 1 dx = lim 1 + x2 t t 1 dx = lim arctan(t) - arctan(1) = - = . 1 + x2 t 2 4 4 1 dx . = 1 + x2 4 5. (10 points) Find the volume of the solid of revolution generated by a 2-rotation around 2 the x-axis of the curve y = , where 1 x 4. x 4 V = 1 y(x) 2 4 dx = 1 1 4 dx = 4 - 2 x x V = 3 . 4 1 = 4 1 - 1 4 = 3. 6. (10 points) Find the area of the region that lies inside the circle r = 1 and outside the curve r = 1 - sin(). Let r1 () = 1 and r2 () = 1 - sin(). Then, -1 y -2 -1 1 r=1 r = 1 - sin(0) A= 1 2 2 r1 () - r2 () d, 2 0 1 2 = 1 - 1 - sin() d, 2 0 1 1 - 1 - sin2 () + 2 sin() d, = 2 0 1 1 - 1 - cos(2) + 2 sin() d, = 2 0 2 1 = -1 + cos(2) + 4 sin() d, 4 0 1 1 = - + sin(2) 0 -4 cos() 0 , 4 2 1 = - - 4(-1 - 1) , 4 =2- . 4 x A =2- . 4 7. (8 points) Find the radius of convergence of the series n x-1 . 2 2 n=0 (8 points) Find the sum of this series for those values of x where the series converges. n-1 |an+1 | (n + 1) |x - 1|n 2 2n-1 |x - 1| n + 1 |x - 1| = = = n n-1 |an | 2 2 n |x - 1| 2 n 2 |an+1 | |x - 1| = <1 n |an | 2 lim |x - 1| < 2. 1+ 1 n Therefore the radius of convergence is R = 2. n x-1 2 2 n=0 n-1 = = = = n=0 n=0 x-1 2 x-1 2 1 (x-1) 2 n , n , , 2 , 3-x 2 = . (3 - x)2 = 2 . (3 - x)2 1- n=0 n x-1 2 2 n-1 8. (10 points) Find the Taylor polynomial of order 3 centered at x = 1 for the function f (x) = ln(x). f (x) = ln(x), 1 f (x) = , x 1 f (x) = - 2 , x 2 f (x) = 3 , x f (1) = 0, f (1) = 1, f (1) = -1, f (1) = 2, T3 (x) = f (1) + f (1) (x - 1) + T3 (x) = (x - 1) - f (1) f (1) (x - 1)2 + (x - 1)3 , 2 6 1 1 (x - 1)2 + (x - 1)3 . 2 3 9. (10 points) (a) Find a solution y(x) of the differential equation (4 + x2 ) y = xy. (b) Find the particular solution y(x) of the above equation that satisfies y(0) = 1. Show all your work. (a) (4 + x2 ) y = xy y x = y 4 + x2 y dx = y x dx. 4 + x2 Substitute u = y(x), then du = y dx, and v = 4 + x2 , then dv = 2x dx. du 1 = u 2 dv v ln(u) = 1 ln(v) + c = ln( v) + c 2 y(x) = 4 + x 2 ec . ln(y) = ln 4 + x2 + c, (b) 1 = y(0) = c 4e ec = 1 2 y(x) = 1 2 4 + x2 . ...
View Full Document

Ask a homework question - tutors are online