Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Definite Integrals anyone?lol

OpenStudy (anonymous):

\[\int\limits_{1}^{e}\left( x- 1/x\right)dx\]

OpenStudy (lalaly):

is it\[\int\limits_{1}^{e}\frac{x-1}{x}dx\]?

OpenStudy (anonymous):

oh sorry no its x - (1/x)

OpenStudy (anonymous):

e^2/2-Ln(e)-1/2+Ln(1)

OpenStudy (lalaly):

\[\int\limits\limits_{1}^{e}(x-\frac{1}{x})dx\]\[\int\limits{x-\frac{1}{x}dx}=\int\limits{xdx}-\int\limits{\frac{1}{x}dx}=\frac{x^2}{2}-\ln|x|\] when u evaluate it from 1 to e\[(\frac{e^2}{2}-\ln(e))-(\frac{1}{2}-\ln(1))\]

OpenStudy (lalaly):

\[=\frac{e^2}{2}-1-\frac{1}{2}-0\]

OpenStudy (anonymous):

see thats what i got but the book has (e^2 - 3)/2 maybe wrong?

OpenStudy (lalaly):

thats the same when u simplify it\[\frac{e^2}{2}-\frac{2}{2}-\frac{1}{2}=\frac{e^2}{2}-\frac{3}{2}=\frac{1}{2}(e^2-3)\]

OpenStudy (anonymous):

ohhh ok thank you lalaly

OpenStudy (lalaly):

:D:D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!