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Improper Integral of

$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$

from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$

The integral, substituting $u= (2x-1)$:

$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$

Definite solution to use with limit:

$$ \frac{1}{2}\cdot\frac{-2}{(2x-1)^2)}\Big|^0_a = \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right)$$

Then solving limit:

$$\lim_{x \to {-\infty}} \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right) = \frac{1}{2}\cdot(-2+0) = -1$$

Any leads?

Thank you.

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You made correct substitution $$u= 2x-1\iff du=2dx$$ $$\int\frac{dx}{(2x-1)^3}dx=\frac12\int\frac{du}{u^3}dx=-\frac{1}{4u^2}+C$$

$$\int\frac{dx}{(2x-1)^3}=-\frac{1}{4(2x-1)^2}+C$$

I hope you can take it from here!

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    $\begingroup$ But, $\frac{1}{2}\int{u^{-3}du} = \frac{1}{2}\frac{u^{-2}}{-2} = \frac{u^{-2}}{-4}$ $\endgroup$ – Fabiano Araujo Nov 22 '14 at 18:33
  • $\begingroup$ if $\frac{u^{-2}}{-4}$ then $\frac{-4}{u^{2}}$ $\endgroup$ – Fabiano Araujo Nov 22 '14 at 18:36
  • $\begingroup$ Ok, how would it result to: $-\frac{1}{4u^2}$, if $\frac{1}{2}\frac{u^{-2}}{2} = \frac{-4}{u^2}$? $\endgroup$ – Fabiano Araujo Nov 22 '14 at 18:40
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Rather than multiplying by $-2,$ you should be dividing, since $$\int u^{n}\,du=\frac1{n+1}u^{n+1}$$ for all integers $n\ne-1.$

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