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Math Help - Integral involving e^(-x^2).

  1. #1
    Newbie penumbra's Avatar
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    Integral involving e^(-x^2).

    I hope that this is the right place to post this.

    I'm studying for my exams this summer and on one of the practice papers there is this integral, which I'm not entirely sure how to approach:

    \int{\exp{(-x^2)}\sqrt{x^4 + 1}\sin{(x)}}dx

    Can anybody help me with this?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by penumbra View Post
    I hope that this is the right place to post this.

    I'm studying for my exams this summer and on one of the practice papers there is this integral, which I'm not entirely sure how to approach:

    \int{\exp{(-x^2)}\sqrt{x^4 + 1}\sin{(x)}}dx

    Can anybody help me with this?

    Thanks in advance!
    was this a originally a definite integral? ... if not, I doubt it can be done.
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  3. #3
    Newbie penumbra's Avatar
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    It was, I'm sorry. It was between pi and -pi.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by penumbra View Post
    It was, I'm sorry. It was between pi and -pi.
    Try doing \int_{-\pi}^{\pi}=\int_{-\pi}^{0}+\int_0^{\pi} and making the sub x=-z in the first integral.
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  5. #5
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    Quote Originally Posted by penumbra View Post
    It was, I'm sorry. It was between pi and -pi.
    easy then. the definite integral's value is 0.

    now ... why?
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  6. #6
    Newbie penumbra's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Try doing \int_{-\pi}^{\pi}=\int_{-\pi}^{0}+\int_0^{\pi} and making the sub x=-z in the first integral.
    Ahh, I see. I think I will be able to do this now. Thank you.
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  7. #7
    Newbie penumbra's Avatar
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    Quote Originally Posted by skeeter View Post
    easy then. the definite integral's value is 0.

    now ... why?
    Because sine of 0, pi and -pi is 0?
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  8. #8
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    Quote Originally Posted by penumbra View Post
    Because sine of 0, pi and -pi is 0?
    no.

    note that the integrand is an odd function.
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  9. #9
    Newbie penumbra's Avatar
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    Quote Originally Posted by skeeter View Post
    no.

    note that the integrand is an odd function.
    So, then \int^{\pi}_0f(x)dx = -\int^0_{-\pi}f(x)dx and so \int^{\pi}_0f(x)dx + \int^0_{-\pi}f(x)dx = 0?
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  10. #10
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    if f(x) is an odd function, then \int_{-a}^a f(x) \, dx = 0

    this is the entire point of the question.
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  11. #11
    Newbie penumbra's Avatar
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    OK, thank you.
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