Partial Fractions Quiz

AttachmentSize
F2010M2300_quiz003_solutions.pdf94.18 KB

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, September 3, 2010

  1. Explain how you know that an integration problem might involve integration by parts. That is, what clues you in to the idea that integration by parts might be a valid approach to solving an integral?

    Solution: The main thing I am looking for here is something along the lines of the following: The integrand appears to be a product of two functions u and v^{{\prime}}, where we can find u^{{\prime}} and v and the integral of u^{{\prime}}v is in some way easier than the original integral. But, there are other instances when integration by parts comes up that don't look like this, such as when integrating \ln x.

  2. Solve the integral, showing all calculations that lead to your result. (You may use a calculator, but your calculations need to be capable of being followed as if the integral is computed by hand.)

    \int\frac{3}{(x-2)(x-5)}\, dx

    Solution: This is an obvious partial fractions problem. First, write

    \int\frac{3}{(x-2)(x-5)}\, dx=3\int\frac{1}{(x-2)(x-5)}\, dx.

    and then compute the partial fractions.

    \frac{1}{(x-2)(x-5)}=\frac{A}{x-2}+\frac{B}{x-5}

    Multiplying on both sides by (x-2)(x-5) and then putting the polynomial on the right in standard form gives

    \displaystyle 1 \displaystyle=A(x-5)+B(x-2)
    \displaystyle 1 \displaystyle=(A+B)x-5A-2B

    This results in the system of equations:

    \displaystyle A+B \displaystyle=0
    \displaystyle-5A-2B \displaystyle=1

    which has the solution A=-\frac{1}{3} and B=\frac{1}{3}. So, we have

    \displaystyle 3\int\frac{1}{(x-2)(x-5)}\, dx \displaystyle=3\int\frac{-1/3}{x-2}+\frac{1/3}{x-5}\, dx
    \displaystyle=-\int\frac{1}{x-2}\, dx+\int\frac{1}{x-5}\, dx
    \displaystyle=-\ln|x-2|+\ln|x-5|+C.

© 2011 Jason B. Hill. All Rights Reserved.