Question

Calculo
Integrais Duplas

$\int\limits^{ln 3}_0 \int\limits^{ln 2}_0 {e^{x+y}} \, dydx \,$

Answer 1

Olá



$\displaystyle \mathsf{ \int\limits^{\ell n (3)}_0 \int\limits^{\ell n (2)}_0 {e^{x+y}} \, dy \, dx }$


Integrando primeiramente em relação a 'y', com isso, 'x' se torna constante.



$\displaystyle \mathsf{ \int\limits^{\ell n (3)}_0 \left[\int\limits^{\ell n (2)}_0 {e^{x+y}} \, dy \right]\, dx }\\\\\\\\\mathsf{ \int\limits^{\ell n (3)}_0 \left[e^{x+y} \frac{}{} \bigg|^{\ell n (2)}_0 \right]\, dx }\\\\\\\\\mathsf{ \int\limits^{\ell n (3)}_0 \left[e^{x+{\ell n (2)}}-e^{x+0} \frac{}{} \right]\, dx }\qquad\qquad\qquad \qquad\boxed{\mathsf{e^{\ell n a}}~=~a}\\\\\\\\\mathsf{ \int\limits^{\ell n (3)}_0 2e^{x}}-e^{x} \, dx }$



Integrando em 'x'



$\displaystyle \mathsf{ \int\limits^{\ell n (3)}_0 2e^{x}-e^{x} \, dx }\\\\\\\\\mathsf{=2e^x-e^x \bigg |^{\ell n (3)}_0\frac{}{} }\\\\\\\\\mathsf{=(2e^{\ell n (3)}-e^{\ell n (3)})-(2e^0-e^0)}\\\\\\\mathsf{=(6-3)-(2-1)}\\\\\\\boxed{\mathsf{=2}}$