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Math Help - Find the integral of a function

  1. #1
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    Find the integral of a function

    I'm working the practice problems in my book, and I keep getting this question wrong. Can you show me the steps of how to solve this, please, because I have no idea of what to do? I changed the numbers, so I could try to solve the question in the book myself. I just want to see the steps.

    ∫(1-x)⁴x dx
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  2. #2
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    Re: Find the integral of a function

    Try rewriting this as \displaystyle -\frac{1}{2}\int{-2x(1 - x^2)^4\,dx}. Does the required substitution become a bit more obvious this way?
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  3. #3
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    Re: Find the integral of a function

    Hello, RedonYellow!

    \int (1-x^2)^4\,x\,dx

    \text{Let: }u \,=\,1-x^2 \quad\Rightarrow\quad du \,=\,\text{-}2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\text{-}\tfrac{1}{2}du

    \text{Substitute: }\;\int u^4\left(\text{-}\tfrac{1}{2}\,du\right) \;=\;\text{-}\tfrac{1}{2}\int u^4\,du

    Got it?

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  4. #4
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    Re: Find the integral of a function

    Of course, you could have noted that (a+ b)^4= a^4+ 4a^3b+ 6a^2b^2+ 4a^3b+ b^4 so that (1- x^2)^4= 1- 4x^2+ 6x^4- 4x^6+ x^8 and then x(1- x^2)^4= x- 4x^3+ 6x^5- 4x^7+ x^9. But, yes, it is easier to use the substitution.
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