What is the definite integral of $e^{2 x}$ $e^(2x)$ from -2 to 2?

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Answer 1 · 2014-09-25T04:23:56

$\int_{- 2}^{2} e^{2 x} d x$ $int_(-2)^(2)e^(2x)dx$

Let $u = 2 x$ $u=2x$

$d u = 2 d x$ $du=2 dx$

$\frac{d u}{2} = \frac{2 d x}{2}$ $(du)/2=(2 dx)/2$

$\frac{d u}{2} = d x$ $(du)/2=dx$

$\int e^{u} \cdot \frac{d u}{2} = \frac{1}{2} \int e^{u} d u$ $inte^(u)*(du)/2=1/2inte^udu$

Calculate the new boundaries

Upper Bound = $2$ $2$

$u = 2 x = 2 (2) = 4$ $u=2x=2(2)=4$

Lower Bound = $- 2$ $-2$

$u = 2 x = 2 (- 2) = - 4$ $u=2x=2(-2)=-4$

$\frac{1}{2} \int_{- 4}^{4} e^{u} d u$ $1/2int_(-4)^(4)e^udu$

$= \frac{1}{2} {[e^{u}]}_{- 4}^{u}$ $=1/2[e^u]_(-4)^(4)$

$= \frac{1}{2} [e^{4} - e^{- 4}]$ $=1/2[e^(4)-e^(-4)]$

$= \frac{1}{2} [e^{4} - \frac{1}{e^{4}}]$ $=1/2[e^(4)-1/e^(4)]$

$= [\frac{e^{4}}{2} - \frac{1}{2 e^{4}}]$ $=[e^(4)/2-1/(2e^(4))]$

$= 27.2899172$ $=27.2899172$

What is the definite integral of e^(2x)e2xe^(2x) from -2 to 2?