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Math Help - A Complicated Integral

  1. #1
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    A Complicated Integral

     d/dx \int_3^{x^2} (2t^2+5)^2 dx

    Can someone please see if I did it right or not?

    x^2=2x

    Substitute t with x^2

    (2(x^2)^2+5)^2

    <br />
2x(2x^4+5)^2

    Do I have to add +C?

    Thanks a lot
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  2. #2
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    Quote Originally Posted by PPSSAGW View Post
     d/dx \int_3^x^2 (2t^2+5)^2 dx

    Do you mean

    \displaystyle\frac{d}{dx}\int_3^{x^2}(2*t^2+5)^2dx
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  3. #3
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    Yeah I later changed it.
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  4. #4
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    Quote Originally Posted by PPSSAGW View Post
    Yeah I later changed it.
    Your x^2 needs to be entered in as ^{x^2} because you have a floating 2.
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  5. #5
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    Thanks. I'm new to LaTex.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Your x^2 needs to be entered in as ^{x^2} because you have a floating 2.
    Did I solve it correctly though? Do I have to add +C?
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  7. #7
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    Quote Originally Posted by PPSSAGW View Post
    Did I solve it correctly though? Do I have to add +C?
    \displaystyle\int_3^{x^2}(2*t^2+5)^2dx\Rightarrow (2*t^2+5)^2\int_3^{x^2} dx=(2*t^2+5)^2\left[x\right]_3^{x^2}=(2*t^2+5)^2[x^2-3]

    No plus C and now differeniate.
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  8. #8
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    There is no need to actually do the integral- this is an exercise in using the "Fundamental Theorem of Calculus":
    \frac{d}{dx}\int_a^x f(t)dt= f(x)
    Here, because the upper limit is x^2 rather than x, you need to use the substitution u= \sqrt{t}= t^{1/2}. That way the upper limit is \sqrt{x^2}= x (assuming x is positive). Now du= (1/2)t^{1/2}dt and t= u^2 so that dt= \frac{2du}{t^{1/2}}= 2\frac{du}{u} and the integral becomes
    2\int_{\sqrt{3}}^x (2u+ 5)^2\left(\frac{du}{u}\right)

    and, by the Fundamental Theorem of Mathematics, the derivative of that is
    2\frac{(2x+ 5)^2}{x}
    just by replacing u in the integrand by x.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    There is no need to actually do the integral- this is an exercise in using the "Fundamental Theorem of Calculus":
    \frac{d}{dx}\int_a^x f(t)dt= f(x)
    Here, because the upper limit is x^2 rather than x, you need to use the substitution u= \sqrt{t}= t^{1/2}. That way the upper limit is \sqrt{x^2}= x (assuming x is positive). Now du= (1/2)t^{1/2}dt and t= u^2 so that dt= \frac{2du}{t^{1/2}}= 2\frac{du}{u} and the integral becomes
    2\int_{\sqrt{3}}^x (2u+ 5)^2\left(\frac{du}{u}\right)

    and, by the Fundamental Theorem of Mathematics, the derivative of that is
    2\frac{(2x+ 5)^2}{x}
    just by replacing u in the integrand by x.
    I thought that too but the integral isn't with respect to t.

    Is it ok to change dx to dt?
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  10. #10
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    Good point but since, unless the intent was
    \frac{d}{dx}\int_{3}{x^2} (2t^2+ 5)^2 dt
    It really is very close to a trivial question!
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    Good point but since, unless the intent was
    \frac{d}{dx}\int_{3}{x^2} (2t^2+ 5)^2 dt
    It really is very close to a trivial question!
    I wanted to do that first but decided to just go what was there. It probably is supposed to be dt.
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