Answer to Question #255372 in Calculus for Aryaf

Question

$\iint$ y/x^2+1da where d = (x,y) 0<x<4 and 0<y< X

Answer 1

$∬ \frac{y}{x^2+1}da \\=∬ \frac{y}{x^2+1} dydx \\=\int_{0}^{4}(\int_{0}^{x}\frac{y}{x^2+1}dy)dx \\=\int_{0}^{4} \frac{1}{2}(\frac{y^2}{x^2+1})^{x}_{0}dx \\=\frac{1}{2}\int_{0}^{4} \frac{x^2}{x^2+1}dx \\=\frac{1}{2}\int_{0}^{4} \frac{x^2+1-1}{x^2+1}dx \\=\frac{1}{2}[\int_{0}^{4} (1-\frac{1}{x^2+1})]dx \\=\frac{1}{2}[\int_{0}^{4} (dx-\frac{1}{x^2+1}dx)] \\=\frac{1}{2}[ x-tan^{-1}x]_{0}^{4} \\=\frac{1}{2}[ 4-tan^{-1}4-0] \\=2-\frac{1}{2}tan^{-1}4$