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Calculus1 14 Online
OpenStudy (anonymous):

evaluate the improper integral from (negative infinity to 0) of 1/(3-4x)dx

OpenStudy (zarkon):

\[\int\limits_{-\infty}^{0}\frac{1}{3-4x}dx\]

OpenStudy (anonymous):

yes, thats it.

OpenStudy (zarkon):

you can either integrate it (you will get a log function) or you can use a comparison test

OpenStudy (anonymous):

i integrated it. I got 1/16Ln(3-4x)

OpenStudy (zarkon):

why 16

OpenStudy (anonymous):

sorry, it souled be (-1/4)ln(3-4x)

OpenStudy (zarkon):

that is better

OpenStudy (anonymous):

im not sure what the new limits of integration are....is it from t to 0?

OpenStudy (zarkon):

\[\int\limits_{-\infty}^{0}\frac{1}{3-4x}dx\] \[\lim_{t\to -\infty}\int\limits_{t}^{0}\frac{1}{3-4x}dx\] \[\lim_{t\to -\infty}\left.-\frac{1}{4}\ln(3-4x)\right|_{t}^{0}\]

OpenStudy (anonymous):

ok that makes sense now. Thanks alot

OpenStudy (zarkon):

np

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