Question

Considere a seguinte integral
$\int\limits^\frac{1}{2} _0 {x} \,e^{2x} \, dx.$
Ela é igual a:

e/4

e/2 + 1

1/2

-1/4

1/4

Answer 1

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$\displaystyle\sf{\int_{0}^{\frac{1}{2}}x~e^{2x}~dx}\\\sf{fac_{\!\!,}a~u=x\implies du=dx}\\\sf{dv=e^{2x}~dx\implies v=\dfrac{1}{2}e^{2x}}\\\displaystyle\sf{\int x~e^{2x}~dx=x\cdot\dfrac{1}{2}e^{2x}-\int \dfrac{1}{2}e^{2x}~dx}\\\displaystyle\sf{\int x~e^{2x}~dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{2}\cdot\dfrac{1}{2}e^{2x}+k}\\\displaystyle\sf{\int x~e^{2x}~dx=\dfrac{1}{2}e^{2x}(x-\dfrac{1}{2})+k}$

$\displaystyle\sf{\int_{0}^{\frac{1}{2}}x~e^{2x}~dx=\dfrac{1}{2}e^{2x}(x-\dfrac{1}{2})\Bigg|_{0}^{\frac{1}{2}}}\\\sf{\dfrac{1}{2}e^{2\cdot\frac{1}{2}}\cdodt(\dfrac{1}{2}-\dfrac{1}{2})-\dfrac{1}{2}e^{2\cdot0}\cdot(0-\dfrac{1}{2})=\dfrac{1}{2}e\cdot0+\dfrac{1}{4}e^0=\dfrac{1}{4}}\\\displaystyle\sf{\int_{0}^{\frac{1}{2}}xe^{2x}~dx=\dfrac{1}{4}\checkmark}$