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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    Evaluate this Integral

    x(x-1)^4 with upper limit 2 and lower limit 1.

    Use the substitution u=x-1

    Answer should be 11/30, but I get no where near that, I think I may be integrating it wrong

    Thanks in advance.
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  2. #2
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    Hello, Hoopy!

    We could find your error if you would show your work . . .


    Evaluate: . \int^2_1x(x-1)^4\,dx

    Let: u = x-1\quad\Rightarrow\quad x = u+1\quad\Rightarrow\quad dx = du

    Substitute: . \int^1_0(u+1)u^4\,du \;=\;\int^1_0\left(u^5 + u^4\right)\,du\;= \;\frac{1}{6}u^6 + \frac{1}{5}u^5\,\bigg]^1_0 \;=\;\frac{11}{30}

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  3. #3
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    Quote Originally Posted by Hoopy View Post
    Evaluate this Integral

    x(x-1)^4 with upper limit 2 and lower limit 1.

    Use the substitution u=x-1

    Answer should be 11/30, but I get no where near that, I think I may be integrating it wrong

    Thanks in advance.
    \int^2_1 x(x-1)^4 \, \mathrm{d}x
    u=x-1 \therefore \frac{\mathrm{d}u}{\mathrm{d}x} = 1 \implies {\mathrm{d}u}={\mathrm{d}x}

    Replace x(x-1)^4 with u and x with u + 1.

    \therefore limits: x=2 \implies u= x-1 = 2-1 =1
    \therefore limits: x=1 \implies u= x-1 = 1-1 =0

    \therefore \int^1_0 (u+1)(u)^4 \, \mathrm{d}u
    = \int^1_0 (u^5 + u^4) \, \mathrm{d}u
    = [\frac{u^6}{6} + \frac{u^5}{5}]^1_0
    =[\frac{11}{30}] - [0] = \frac{11}{30}
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  4. #4
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    Oops, A minute late.
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  5. #5
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    Cheers guys, I just forgot to change everything to u.
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