1. ## antiderivative

(integral symbol) (7e^(2x)-3e^-x) dx

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

2. $\displaystyle \int e^{ax} dx = \frac{e^{ax}}{a}+c$

3. 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

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

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

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

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

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

Therefore (subbing back for u)

$\displaystyle =\frac{7}{2}e^{2x}+3e^{-x}+C$

4. Originally Posted by jwbehm
thanks.
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