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Thread: Integral x^4(4+x^5)^1/2

  1. #1
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    Integral x^4(4+x^5)^1/2

    $\displaystyle \int x^4\sqrt{4+x^5}$

    Do I make $\displaystyle u=4+x^5$ in order to solve this problem?
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by cammywhite View Post
    $\displaystyle \int x^4\sqrt{4+x^5}$

    Do I make $\displaystyle u=4+x^5$ in order to solve this problem?
    Yes! Just remember to factor $\displaystyle \frac{1}{5}$ out of your integral!
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  3. #3
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    Quote Originally Posted by cammywhite View Post
    $\displaystyle \int x^4\sqrt{4+x^5}$

    Do I make $\displaystyle u=4+x^5$ in order to solve this problem?
    Yes.

    Notice that if $\displaystyle u = 4 + x^5$ then $\displaystyle \frac{du}{dx} = 5x^4$.


    So $\displaystyle \int{x^4\sqrt{4 + x^5}\,dx} = \frac{1}{5}\int{5x^4(4 + x^5)^{\frac{1}{2}}\,dx}$

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx }$

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$.


    I trust you can go from here...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Yes.

    Notice that if $\displaystyle u = 4 + x^5$ then $\displaystyle \frac{du}{dx} = 5x^4$.


    So $\displaystyle \int{x^4\sqrt{4 + x^5}\,dx} = \frac{1}{5}\int{5x^4(4 + x^5)^{\frac{1}{2}}\,dx}$

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx }$

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$.


    I trust you can go from here...

    Let me try

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
    $\displaystyle = \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

    so far correct?
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by cammywhite View Post
    Let me try

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
    $\displaystyle = \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

    so far correct?
    Your integration is correct, but you need to leave out the integral sign & du since you have already integrated.
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  6. #6
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    Quote Originally Posted by mollymcf2009 View Post
    Your integration is correct, but you need to leave out the integral sign & du since you have already integrated.
    I looked at the answer and it's $\displaystyle \frac {\sqrt{x^5+4}(2x^5+8)}{15}$

    I have no clue how to get to the answer
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by cammywhite View Post
    Let me try

    $\displaystyle = \frac{1}{5}\int{u^{\frac{1}{2}}\,du}$
    $\displaystyle = \frac{1}{5}\int\frac{{u^{\frac{3}{2}}\,du}}{\frac{ 3}{2}}$

    so far correct?
    $\displaystyle
    \frac{1}{5} \left( \frac{u^{3/2}}{\frac{3}{2}}\right)=\frac{2}{15}u^{3/2}
    $

    $\displaystyle \frac{2}{15}(4+x^5)^{3/2}=\frac{2}{15}\sqrt{4+x^5}(4+x^5)=\frac{2(4+x^5)\ sqrt{4+x^5}}{15}=\frac{(8+2x^5)\sqrt{4+x^5}}{15}$
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  8. #8
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    Quote Originally Posted by TheEmptySet View Post
    $\displaystyle
    \frac{1}{5} \left( \frac{u^{3/2}}{\frac{3}{2}}\right)=\frac{2}{15}u^{3/2}
    $

    $\displaystyle \frac{2}{15}(4+x^5)^{3/2}=\frac{2}{15}\sqrt{4+x^5}(4+x^5)=\frac{2(4+x^5)\ sqrt{4+x^5}}{15}=\frac{(8+2x^5)\sqrt{4+x^5}}{15}$

    i don't get this part

    $\displaystyle \frac{2}{15}(4+x^5)^{3/2} =\frac{2}{15}\sqrt{4+x^5}(4+x^5)$
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  9. #9
    Behold, the power of SARDINES!
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    Quote Originally Posted by cammywhite View Post
    i don't get this part

    $\displaystyle \frac{2}{15}(4+x^5)^{3/2} =\frac{2}{15}\sqrt{4+x^5}(4+x^5)$
    We are using this exponent law

    $\displaystyle x^ax^b=x^{a+b}$

    Note that $\displaystyle x^{\frac{3}{2}}=x^{\frac{1}{2}}x^1=\sqrt{x}\cdot x$

    So using this we get

    $\displaystyle (4+x^5)^{3/2} =\sqrt{4+x^5}(4+x^5)$
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  10. #10
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    Integration by substitution is like a backward-application of the chain rule. In the expression

    $\displaystyle x^4\sqrt{4+x^5},$

    we see that $\displaystyle x^4$ looks like it was differentiated from $\displaystyle 4+x^5$ of the square root.

    To find the real antiderivative, we just add in the factor $\displaystyle \frac{1}{5}$ to get $\displaystyle x^4$ from $\displaystyle 5x^4$.
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