Irrerbthist6n

2021-12-13

Tell, what is the antiderivative of ${e}^{2x}$?

Bernard Lacey

$\int {e}^{2x}=1\frac{l}{2}\int 2{e}^{2x}dx$
If $2dx=d\left(2x\right)$, 2 is the derivative of 2x
And we have:
$\frac{1}{2}\int {e}^{2x}d\left(2x\right)$
Let u=2x
Thus, $\frac{1}{2}\int {e}^{u}du=\frac{1}{2}{e}^{u}=\frac{1}{2}{e}^{2x}$

Annie Levasseur

You can use the technique of integration by substitution to find this.
$f\left(x\right)=\frac{1}{2}{e}^{2x}$ has ${f}^{\prime }\left(x\right)={e}^{2x}$, so it's an antiderivative. And the general one is:
$\frac{1}{2}{e}^{2x}+C$

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