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Math Help - Definite Integral

  1. #1
    Junior Member Hardwork's Avatar
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    Definite Integral

    Find \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Is that what you are looking for?

    \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}\times \sqrt{\sqrt{15}+2x}}{\sqrt{(2-x)(2+x)}(4-x^2)^2}\ dx

    I only simplified the brackets and replace an 'x' by 'times', and simplified the denominator a little and added dx.

    As you can see, the denominator and numerator can be further simplified.

    \displaystyle \int_{-1}^{1}\frac{\sqrt{(\sqrt{15}-2x)(\sqrt{15}+2x)}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx

    \displaystyle \int_{-1}^{1}\frac{\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx

    Can you continue now?

    EDIT: Since the 'x' is an actual x, this becomes:

    \displaystyle \int_{-1}^{1}\frac{x\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx
    Last edited by Unknown008; October 30th 2010 at 09:57 AM.
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  3. #3
    A Plied Mathematician
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    \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.

    Do you notice anything about the integrand?
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  4. #4
    Junior Member Hardwork's Avatar
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    @Unknown - It was an actual x. I'm sorry I should have made it clear.
    Quote Originally Posted by Ackbeet View Post
    \displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.

    Do you notice anything about the integrand?
    Looks reasonable now. I'll try to take it from there.
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  5. #5
    A Plied Mathematician
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    You should be able to write the answer down right now, with no further computation.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, it just seemed strange that you'd have an x in the middle like that =P
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  7. #7
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Ackbeet View Post
    You should be able to write the answer down right now, with no further computation.
    You mean I don't need to find the antiderivative? That would be amazing, but I'm not sure how to do it. Is it to do with the domains of \sqrt{15-4x^2} and \sqrt{(4-x^2)^5}?
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  8. #8
    A Plied Mathematician
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    You mean I don't need to find the antiderivative?
    Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?
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  9. #9
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?
    Limits flip and the sign changes - i.e. we get \displaystyle -\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.<br />
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  10. #10
    A Plied Mathematician
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    Actually, the limits flip, but the overall sign does NOT change. That's because you get one minus sign for the differential, and one minus sign for the lone x in the numerator. So you get

    \displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx=\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.

    Now, what happens when you flip the right integral's limits back to what they are on the LHS?
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  11. #11
    Junior Member Hardwork's Avatar
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    You are right. I forgot to account for the minus from the differential.
    Quote Originally Posted by Ackbeet View Post
    Now, what happens when you flip the right integral's limits back to what they are on the LHS?
    The integral on the right is

    \int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx

    Flipping the limits to what they were

    \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx

    I get the integral in the left.
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  12. #12
    A Plied Mathematician
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    No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).
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  13. #13
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Ackbeet View Post
    No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).
    Oh, I knew I was missing something.

    So \int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx

    So \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx
    Last edited by Hardwork; October 30th 2010 at 12:13 PM.
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  14. #14
    A Plied Mathematician
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    So, if y = -y, what is y?
    Last edited by Ackbeet; October 30th 2010 at 12:26 PM. Reason: Different variable.
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  15. #15
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Ackbeet View Post
    So, if y = -y, what is y?
    0

    That's it?
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