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Math Help - Difficult definite integral

  1. #1
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    Difficult definite integral

    The question:
    \int_{-1}^{0} \sqrt{t^2 + t^4} dt

    I'm stumped. Not sure how to attempt this.

    Any assistance would be great.
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  2. #2
    Member HappyJoe's Avatar
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    How about first taking t^2 from within the square root, and move it outside as |t|:

    \sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},

    where |t|=-t, because we find ourselves in the range -1\leq t\leq 0. This last expression can be more easily integrated. I suggest substitution.
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  3. #3
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    Quote Originally Posted by Glitch View Post
    The question:
    \int_{-1}^{0} \sqrt{t^2 + t^4} dt

    I'm stumped. Not sure how to attempt this.

    Any assistance would be great.

    The integrand function is even, so we can as well work with \int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li  mits^1_0 2t\sqrt{1+t^2}dt = ....take it from here.

    Note: the last one is an automatic integral!

    Tonio
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tonio View Post
    The integrand function is even, so we can as well work with \int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li  mits^1_0 2t\sqrt{1+t^2}dt = ....take it from here.

    Note: the last one is an automatic integral!

    Tonio
    I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

    \displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}

    It is quite obvious that the second term of the 'identity' (1) is an odd function ...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by HappyJoe View Post
    How about first taking t^2 from within the square root, and move it outside as |t|:

    \sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},

    where |t|=-t, because we find ourselves in the range -1\leq t\leq 0. This last expression can be more easily integrated. I suggest substitution.
    I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
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  6. #6
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    Quote Originally Posted by chisigma View Post
    I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

    \displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}

    It is quite obvious that the second term of the 'identity' (1) is an odd function ...

    Kind regards


    \chi \sigma
    This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.

    In fact I should have writtent \sqrt{t^2+t^4}=|t|\sqrt{1+t^2} , but in this case is irrelevant since the

    variable t is already positive in the new integration's limits.

    Tonio
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  7. #7
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    Quote Originally Posted by Glitch View Post
    I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
    Let -u = t, then dt = -du, then:

    \displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}.

    Now we can write \sqrt{t^2+t^4} = t\sqrt{1+t^2}, as our limits are non-negative.
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