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Thread: Evaluate the integral

  1. #1
    Super Member Anonymous1's Avatar
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    Evaluate the integral

    Where k is a positive integer, evaluate:

    $\displaystyle \int_{-1}^{1} \! (1 - x^2)^k \, dx.$

    Possibly make a change of variables? Any help?

    ~Thanks
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  2. #2
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    Quote Originally Posted by Anonymous1 View Post
    Where k is a positive integer, evaluate:

    $\displaystyle \int_{-1}^{1} \! (1 - x^2)^k \, dx.$

    Possibly make a change of variables? Any help?

    ~Thanks
    You can expand it using the Binomial Theorem...
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    Sorry, wrong answer.
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    Quote Originally Posted by chengbin View Post
    Look closer, the answer is 0 no matter what value of k.

    $\displaystyle =\bigg [ \frac{1}{2k+2}(1-x^2)^{k+1} \bigg ]_{-1}^1$

    $\displaystyle =\bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg ) - \bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg )$

    $\displaystyle =0$
    It can't be zero ever, because $\displaystyle (1-x^2)^k$ is an even function on a symmetric interval, so $\displaystyle \int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$. Since $\displaystyle 0\leq1-x^2\leq1$ on $\displaystyle [0,1]$, this integral is always strictly positive.

    Furthermore, the antiderivative is not $\displaystyle \frac{1}{2k+2}(1-x^2)^{k+1}$. The derivative of that function is $\displaystyle -x(1-x^2)^k$.
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    Quote Originally Posted by redsoxfan325 View Post
    It can't be zero ever, because $\displaystyle (1-x^2)^k$ is an even function on a symmetric interval, so $\displaystyle \int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$. Since $\displaystyle 0\leq1-x^2\leq1$ on $\displaystyle [0,1]$, this integral is always strictly positive.

    Furthermore, the antiderivative is not $\displaystyle \frac{1}{2k+2}(1-x^2)^{k+1}$. The derivative of that function is $\displaystyle -x(1-x^2)^k$.
    I realized that after I posted.
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    Super Member redsoxfan325's Avatar
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    I realized that you realized that after I posted.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    You can expand it using the Binomial Theorem...
    Really? The whole $\displaystyle x^2$ thing isn't an issue?
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  8. #8
    Super Member redsoxfan325's Avatar
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    Nope. The binomial theorem works with anything.
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  9. #9
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    Quote Originally Posted by Anonymous1 View Post
    Where k is a positive integer, evaluate:

    $\displaystyle \int_{-1}^{1} \! (1 - x^2)^k \, dx.$

    Possibly make a change of variables? Any help?

    ~Thanks
    Use the substitution $\displaystyle x = \sin{\theta}$ so that $\displaystyle dx = \cos{\theta}\,d\theta$.


    Also note that $\displaystyle 1 - x^2 = 1 - \sin^2{\theta} = \cos^2{\theta}$.


    So $\displaystyle \int{(1 - x^2)^k\,dx} = \int{\cos^{2k}{\theta}\,\cos{\theta}\,d\theta}$

    $\displaystyle = \int{\cos^{2k + 1}{\theta}\,d\theta}$.
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