Question

qual é a integral de e neperiano elevado a -x/2 no intervalo de 3 e -2. Resposta é 4,9904. Alguem me ajuda pfv!

Answer 1

$\displaystyle\mathsf{\int_{-2}^{3} e^{-x/2}\,dx}\\\\\\ =\mathsf{-2\int_{-2}^{3} \left(-\dfrac{1}{2}\right)\cdot e^{-x/2}\,dx}\\\\\\ =\mathsf{-2\int_{-2}^{3} e^{-x/2}\cdot \left(-\dfrac{1}{2}\right)\!dx}~~~~~~\mathbf{(i)}$

Fazendo a seguinte mudança de variável:

$\mathsf{-\dfrac{x}{2}=u~~\Rightarrow~~-\dfrac{1}{2}\,dx=du}$


Mudando os extremos de integração:

$\mathsf{Quando~~x=-2~~\Rightarrow~~u=1}\\\\ \mathsf{Quando~~x=3~~\Rightarrow~~u=-\dfrac{3}{2}}$


Substituindo em $\mathbf{(i)},$ a integral fica

$=\displaystyle\mathsf{-2\int_{1}^{-3/2} e^{u}\,du}\\\\\\ =\mathsf{-2\cdot \left.\left(e^{u} \right )\right|_{1}^{-3/2}}\\\\\\ =\mathsf{-2\cdot (e^{-3/2}-e^{1})}\\\\ =\mathsf{-2e^{-3/2}+2e}\\\\ =\mathsf{-\dfrac{2}{e^{3/2}}+2e\approx 4,9903}$