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Math Help - Intergration problem, absolute value

  1. #1
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    Intergration problem, absolute value

    This is the problem:

    \int_{0}^{2}|x^{2}-x|

    Solution:

    |x^2-x| = x|x-1|

    \int_{0}^{2}|x^{2}-x| = \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1)

    =/ \\ \int  x(x-1) = x^2/2(x-1) - x^3/6 = x^3/2 - x^2/2 - x^3/6 \\ /\\ = [x^3/2 - x^2/2 - x^3/6]_{0}^{1} + [x^3/2 - x^2/2 - x^3/6]^{2}_{1} = 1

    This is the correct answer, however I was wondering why:

    \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1) \neq  \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x

    I got \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x = 2/3 , am I right?
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  2. #2
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    Re: Intergration problem, absolute value

    Quote Originally Posted by Pettzon View Post
    This is the problem:

    \int_{0}^{2}|x^{2}-x|

    Solution:

    |x^2-x| = x|x-1|

    \int_{0}^{2}|x^{2}-x| = \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1)

    =/ \\ \int  x(x-1) = x^2/2(x-1) - x^3/6 = x^3/2 - x^2/2 - x^3/6 \\ /\\ = [x^3/2 - x^2/2 - x^3/6]_{0}^{1} + [x^3/2 - x^2/2 - x^3/6]^{2}_{1} = 1

    This is the correct answer, however I was wondering why:

    \int_{0}^{1} x(x-1) + \int_{1}^{2} x(x-1) \neq  \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x

    I got \int_{0}^{1} x^2-x + \int_{1}^{2} x^2-x = 2/3 , am I right?
    For starters, I think you'll find that \displaystyle |x^2 - x| = |x||x - 1|, not \displaystyle x|x-1|...
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  3. #3
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    Re: Intergration problem, absolute value

    Quote Originally Posted by Pettzon View Post
    This is the problem:

    \int_{0}^{2}|x^{2}-x|
    \int_{0}^{2}|x^{2}-x|=\int_{0}^{1}(x-x^2)+\int_{1}^{2}(x^2-x)
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    Re: Intergration problem, absolute value

    Quote Originally Posted by Prove It View Post
    For starters, I think you'll find that \displaystyle |x^2 - x| = |x||x - 1|, not \displaystyle x|x-1|...
     |x^2 - x| = |x(x-1)| = x|x-1|?


    Quote Originally Posted by Plato View Post
    \int_{0}^{2}|x^{2}-x|=\int_{0}^{1}(x-x^2)+\int_{1}^{2}(x^2-x)
    Of course, thank you!
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  5. #5
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    Re: Intergration problem, absolute value

    Quote Originally Posted by Pettzon View Post
     |x^2 - x| = |x(x-1)| = x|x-1|?




    Of course, thank you!
    Wrong, what if \displaystyle x < 0? It's impossible for \displaystyle |x(x - 1)| to be negative, but \displaystyle x|x - 1| would be.

    The general rule is \displaystyle |AB| = |A||B|, so \displaystyle |x(x-1)| = |x||x-1|.

    Anyway, for this integral, you should know that

    \displaystyle \begin{align*}|x^2 - x| &= \begin{cases}x^2 - x\textrm{ if }x^2 - x \geq 0\\x - x^2\textrm{ if }x^2-x<0 \end{cases} \\ &= \begin{cases}x^2 - x\textrm{ if } x \leq 0 \textrm{ or }x \geq 1 \\ x - x^2 \textrm{ if }0 < x < 1\end{cases} \end{align*}

    so that means

    \displaystyle \begin{align*} \int_0^2{|x^2-x|\,dx} &= \int_0^1{|x^2-x|\,dx} + \int_1^2{|x^2-x|\,dx} \\ &= \int_0^1{x - x^2\,dx} + \int_1^2{x^2 - x\,dx} \end{align*}
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    Re: Intergration problem, absolute value

    Quote Originally Posted by Prove It View Post
    For starters, I think you'll find that \displaystyle |x^2 - x| = |x||x - 1|, not \displaystyle x|x-1|...
    Quote Originally Posted by Prove It View Post
    Wrong, what if \displaystyle x < 0? It's impossible for \displaystyle |x(x - 1)| to be negative, but \displaystyle x|x - 1| would be.

    The general rule is \displaystyle |AB| = |A||B|, so \displaystyle |x(x-1)| = |x||x-1|.

    Anyway, for this integral, you should know that

    \displaystyle \begin{align*}|x^2 - x| &= \begin{cases}x^2 - x\textrm{ if }x^2 - x \geq 0\\x - x^2\textrm{ if }x^2-x<0 \end{cases} \\ &= \begin{cases}x^2 - x\textrm{ if } x \leq 0 \textrm{ or }x \geq 1 \\ x - x^2 \textrm{ if }0 < x < 1\end{cases} \end{align*}

    so that means

    \displaystyle \begin{align*} \int_0^2{|x^2-x|\,dx} &= \int_0^1{|x^2-x|\,dx} + \int_1^2{|x^2-x|\,dx} \\ &= \int_0^1{x - x^2\,dx} + \int_1^2{x^2 - x\,dx} \end{align*}
    Ah, I see. Thanks a bunch!
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