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anonymous

  • 5 years ago

integrate (x)(x^2-1)dx by subst without first multiplying. Why not?

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  1. anonymous
    • 5 years ago
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    Just to get the practice of using a u-substitution; you COULD do it by multiplying, but that's seemingly not the point.

  2. Lost
    • 5 years ago
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    \[\int\limits_{-\infty}^{\infty}(x ^{3}-x)dx\] Start by separating the terms along with the differential: \[\int\limits_{-\infty}^{\infty}x ^{3}dx - \int\limits_{-\infty}^{\infty}xdx\] Integrate both parts: \[(\frac{1}{4}x^4-\frac{1}{2}x^2+c)|^{\infty}_{-\infty}\] I'm posting the above because I just took all the time to make the equations look nice in the editor, but I know realize you were not supposed to multiply first. In this case, you were asked to integrate as given to practice u-substitution as stated earlier. \[u = x^2-1 \] \[du = 2xdx\] \[\frac{1}{2}du=xdx\] \[\frac{1}{2}\int\limits_{-\infty}^{\infty}udu\] \[(\frac{1}{2})(\frac{1}{2})u^2=\frac{1}{4}u^2+c\] Then substitute back in the value of u: \[=\frac{1}{4}(x^2-1)+c\]

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