Algebra and Partial Fractions, Summaries of Algebra

Download Algebra and Partial Fractions and more Summaries Algebra in PDF only on Docsity! Algebra and Partial Fractions - Solutions Math 125 Integration of rational functions is mostly a matter of algebraic manipulation. In this worksheet we shall work through some examples of the necessary techniques. 1a Consider the rational function f(x) = 2x3 − 4x2 − 5x + 3 x2 − 2x − 3 . Use long division to get a quotient and a remainder, then write f(x) =quotient + (remainder/divisor). f(x) = 2x + x + 3 x2 − 2x − 3 1b Now consider the expression x + 3 x2 − 2x − 3 . Factor the denominator into two linear terms. x + 3 (x + 1)(x − 3) 1c We wish to write x + 3 (x − 3)(x + 1) as a sum A x − 3 + B x + 1 . Let’s find A and B. Set the two expressions equal and clear denominators (that is, multiply through by (x − 3)(x + 1) and cancel (x − 3)’s and (x + 1)’s as much as possible). Plug in x = 3 and solve for A. Use the same idea to find B. Check your work by adding the two fractions together. x + 3 = A(x + 1) + B(x − 3) If x = 3, we get 6 = 4A so A = 3 2 . If x = −1, we get 2 = −4B so B = − 1 2 . Check that 3/2 x − 3 + −1/2 x + 1 = x + 3 (x − 3)(x + 1) . 1d Now use the results of Problems 1a, 1b, and 1c to compute ! 2x3 − 4x2 − 5x + 3 x2 − 2x − 3 dx. Putting it all together, f(x) = 2x + 3/2 x − 3 − 1/2 x + 1 so ! f(x) dx = ! 2x + 3/2 x − 3 − 1/2 x + 1 dx = x2 + 3 2 ln |x − 3|− 1 2 ln |x + 1| + C. 1e Some of the terms in the answer to Problem 1d involve logarithms. Combine those terms into a single term of the form ln(some function of x). ! f(x) dx = x2 + ln " # # $ % % % % % (x − 3)3 x + 1 % % % % % + C 3 The regions A and B in the figure are re- volved around the x-axis to form two solids of revolution. (a) Before computing the integrals, which solid do you think has a larger volume? Why? Region B looks larger. A y=ln(x) 1 3 4 51 2 e B (b) Use the disk method to find the volume of the solid swept out by region A. ! e 1 π ln(x)2 dx u = ln(x)2 dv = π dx du = 2 ln(x) x dx v = πx ! π ln(x)2 dx = πx ln(x)2 − ! 2π ln(x) dx = (∗) U = ln(x) dV = 2π dx dU = 1 x dx V = 2πx (∗) = πx ln(x)2 − 2πx ln(x) + ! 2π dx = πx ln(x)2 − 2πx ln(x) + 2πx + C The volume is " πx ln(x)2 − 2πx ln(x) + 2πx #$$$ e 1 = π(e − 2) ≈ 2.2565 (c) Use the shell method to find the volume of the solid swept out by region B. ! 1 0 2πyey dy u = 2πy dv = ey dy du = 2π dy v = ey ! 2πyey dy = 2πyey − ! 2πey dy = 2πyey − 2πey + C The volume is (2πyey − 2πey)|10 = 2π ≈ 6.283