# antiderivative

• July 22nd 2009, 06:14 PM
jwbehm
antiderivative
(integral symbol) (7e^(2x)-3e^-x) dx

Thanks guys i dont know how to do it with the e's
• July 22nd 2009, 06:35 PM
pickslides
$\int e^{ax} dx = \frac{e^{ax}}{a}+c$
• July 22nd 2009, 06:43 PM
VonNemo19
Quote:

Originally Posted by jwbehm
(integral symbol) (7e^(2x)-3e^-x) dx

Thanks guys i dont know how to do it with the e's

Break it up first

$7\int{e^{2x}}-3\int{e}^{-x}$

In the first let $u_1=2x$, then $\frac{1}{2}du_1=dx$

In the second let $u_2=-x$, then $-du_2=dx$
(Iwrite subscripts to avoid confusion, I should have used u AND v, but too late now)...
and we have

$\frac{7}{2}\int{e^{u_1}}du_1+3\int{e^{u_2}}du_2$

The integral of $e^u:e^u$

Therefore (subbing back for u)

$=\frac{7}{2}e^{2x}+3e^{-x}+C$
• July 23rd 2009, 12:33 PM
Chris L T521
Quote:

Originally Posted by jwbehm
thanks.

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