Question

Calculo 2 ayudaaaaaaaaaa

Answer 1

$\int\limits^1_0 {\frac{y}{e^{2y}}} \, dy =-\frac{1}{2}y\,e^{-2y}\Bigg|_0^1+\frac{1}{2}\int\limits^1_0 {e^{-2y}}\\\\=\Big(-\frac{1}{2}y\,e^{-2y}-\frac{1}{4}e^{-2y}} \Big)\Bigg|_0^1\\\\=e^{-2y}\Big(-\frac{1}{2}y-\frac{1}{4}} \Big)\Bigg|_0^1\\\\=e^{-2}\Big(-\frac{1}{2}-\frac{1}{4}} \Big)-e^0\Big(-\frac{1}{4}} \Big)\\\\=\frac{1}{4}-\frac{3}{4}e^{-2}=\boxed{\frac{1}{4}-\frac{3}{4e^2}}$