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Thread: Approximating definite integral using series

  1. #1
    Senior Member mollymcf2009's Avatar
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    Approximating definite integral using series

    Use series to approximate the definite integral I. (correct to 3 decimal places)

    I = \int^{1}_{0} x cos(x^5) dx


    Did I do my series right?

    \sum^{\infty}_{n=0} x (-1)^n \frac{(x)^{10n}}{(2n)!}
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  2. #2
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    Quote Originally Posted by mollymcf2009 View Post
    Use series to approximate the definite integral I. (correct to 3 decimal places)

    I = \int^{1}_{0} x cos(x^5) dx


    Did I do my series right?

    \sum^{\infty}_{n=0} x (-1)^n \frac{(x)^{10n}}{(2n)!}
    \cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...

    \cos(x^5) = 1 - \frac{x^{10}}{2!} + \frac{x^{20}}{4!} - \frac{x^{30}}{6!} + ...

    x\cos(x^5) = x - \frac{x^{11}}{2!} + \frac{x^{21}}{4!} - \frac{x^{31}}{6!} + ...

    integrate ...

    \int_0^1 x\cos(x^5) \, dx = \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{32}}{32 \cdot 6!} + ... \right]_0^1<br />

    to be accurate to 3 decimal places ...

    \int_0^1 x\cos(x^5) \, dx \approx \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!}\right]_0^1 = 0.460<br />
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by skeeter View Post
    \cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...

    \cos(x^5) = 1 - \frac{x^{10}}{2!} + \frac{x^{20}}{4!} - \frac{x^{30}}{6!} + ...

    x\cos(x^5) = x - \frac{x^{11}}{2!} + \frac{x^{21}}{4!} - \frac{x^{31}}{6!} + ...

    integrate ...

    \int_0^1 x\cos(x^5) \, dx = \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{32}}{32 \cdot 6!} + ... \right]_0^1<br />

    to be accurate to 3 decimal places ...

    \int_0^1 x\cos(x^5) \, dx \approx \left[\frac{x^2}{2} - \frac{x^{12}}{12 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!}\right]_0^1 = 0.460<br />
    Awesome! I got exactly what you did, EXCEPT I forgot to add in the extra x before I integrated, so when I integrated, I was getting my exponents one less than I should have.

    Thanks Skeeter!!
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