Question

Resolva a integral dupla:$\int\limits^4_0 { \int\limits^ \frac{y}{4} _0 {e ^{-y ^{2} } } \, dx } \, dy$

Answer 1

Oi. 
Eu arriscaria dessa forma :)


$\int\limits^{ \frac{y}{4} }_0 {e^{-y^2}} \, dx \\ \\ {e^{-y^2}} \int\limits^{ \frac{y}{4} }_0 \, dx \\ \\ {e^{-y^2}} .x \ |^{ \frac{y}{4} }_0 \\ \\ e^{-y^2}( \frac{y}{4} -0) \\ \\ \boxed{e^{-y^2}.\frac{y}{4} }$

Resolvendo em relação a dy com o resultado anterior:

$\int\limits^4_0 {e^{-y^2}.\frac{y}{4}} \, dy \\ \\ -------------- \\ \boxed{ u=-y^2} \\ \\ \frac{du}{dy} =-2y \ \ \ -\ \textgreater \ \boxed{dy= -\frac{du}{2y}} \\ -------------- \\ \\ \int\limits^4_0 {e^{u}.\frac{y}{4}} \, (- \frac{du}{2y} ) \\ \\ - \frac{1}{8} \int\limits^4_0 {e^{u}.du} \\ \\ - \frac{1}{8}(e^u)\ |^4_0 \\ \\ - \frac{1}{8}(e^{-y^2})\ |^4_0 \\ \\ - \frac{1}{8}(e^{-4^2}-e^{-0^2})\ \\ \\ - \frac{1}{8}(e^{-16}-1) \\ \\ - \frac{1}{8}( \frac{1}{e^{16}} -1)$

$\boxed{ - \frac{1}{8e^{16}}+ \frac{1}{8} }$