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Math Help - U Substitution

  1. #1
    Member alexgeek's Avatar
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    Question U Substitution

    Could someone check if I have the right answer for this? I tried wolfram alpha but it used a long division method instead of u substitution.
    <br />
\begin{array}{lll}<br />
\int x(x-3)^2 \: dx <br />
& = & \int (u+3)u^2 \: du \\<br />
& = & \int u^3 + 3u^2 \: du \\<br />
& = & \frac{u^4}{4} + u^2 + c \\<br />
& = & \frac{(x-3)^4}{4} + (x-3)^2 + c<br />
\end{array}<br />
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  2. #2
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    @ alexgeek

    You indeed have got the right answer. It is integration by simple substitution. No offences but I wonder why do you even need to check this on the forum ?
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  3. #3
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    Presumably, you let u= x- 3 so that du= dx and u+ 3= x. That gives, of course,
    \int x(x- 3)^2dx= \int(u+3)u^2 du as you have.

    I have no idea what you mean by "a long division method". You could, of course, do this by multiplying: (x- 3)^2= x^2- 6x+ 9 so \int x(x- 3)^2 dx= \int x^3- 6x^2+ 9x dx.
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  4. #4
    Member alexgeek's Avatar
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    I wanted to check because it looks nothing like the answer wolframalpha produced, so thought it may be wrong.
    Cheers.
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