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OpenStudy (anonymous):

integral rule for e^u?

OpenStudy (anonymous):

\[\int\limits_{?}^{?} e ^{u}\]

OpenStudy (anonymous):

The integral of e^u is e^u + c

OpenStudy (anonymous):

I got that from my calculus book

OpenStudy (anonymous):

is it e^u*u'?

OpenStudy (anonymous):

which one is tht?

OpenStudy (anonymous):

the integral is e^u

OpenStudy (anonymous):

so e^(1-x) is what?

OpenStudy (anonymous):

\[\int\limits_{?}^{?}e^u = e^u + C\]

OpenStudy (anonymous):

\[\int\limits e^u du= e^u\]

OpenStudy (anonymous):

yes +C integration constant

OpenStudy (anonymous):

what is the integral of \[5^{-x}\]

OpenStudy (anonymous):

-5^-x/ln5 +C

OpenStudy (anonymous):

watz the deriv of e^u>?

OpenStudy (anonymous):

Example: If you want to solve \[\int\limits_{?}^{?}e ^{3x+1}dx\] then you rewrite it by doing a substitution with u\[\int\limits_{}^{?}e^udu\], when u=3x+1, then du=3, therefore you write this as:\[(1/3)\int\limits_{?}^{?}e^udu\]=\[(1/3)e^u = (1/3)e ^{3x+1} + C\]

OpenStudy (anonymous):

sorry, du=3dx

OpenStudy (anonymous):

ohhok got it thankssss:)

OpenStudy (anonymous):

integral rule for ln u?

OpenStudy (anonymous):

You will need to integrate by parts to solve that one.

OpenStudy (anonymous):

\[\int\limits_{?}^{?}udv = uv - \int\limits_{?}^{?}vdu\]

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