Friday Quiz 1

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MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, August 27, 2009

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

    Solution: I will accept answers that make sense here. What I would write is as follows. The biggest hint that an integral might involve substitution is when the integral contains an inside function and that inside function's derivative (up to some constant) is somewhere in the outside of the integrand.

  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 _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx

    Solution: There are several ways to solve this. We know what the curve \sin x looks like, and integrating it from -n to n for any real number n will give us zero. If we scale that curve by \pi vertically and compress horizontally by a factor of \pi^{2}, we expect the same to be true. So, right away, we know that we'll end with zero. Using the substitution w=\pi^{2}x we find the following.

    \displaystyle\int _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx \displaystyle=\frac{1}{2}\int _{{-\pi^{3}}}^{{\pi^{3}}}\sin(w)\, dw
    \displaystyle=\frac{1}{2}\left[-\cos w\right]_{{-\pi^{3}}}^{{\pi^{3}}}
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(-\pi^{3})\right]
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(\pi^{3})\right]
    \displaystyle=0.

    The second to last step is true because \cos(\theta)=\cos(-\theta) for all \theta.

© 2011 Jason B. Hill. All Rights Reserved.