## Calculus with Applications (10th Edition)

$=\frac{1}{2}\ln5$
We are given $\int^{2}_{0}\int^{1}_{1/2}\frac{1}{y^{2}+1}dydx=\int^{2}_{0}(\int^{1}_{1/2}\frac{1}{y^{2}+1}dy)dx$ solve the inner integral $\int^{1}_{1/2}\frac{1}{y^{2}+1}dy$ $=\frac{1}{y^{2}+1}\int^{1}_{1/2}dy$ $=\frac{1}{y^{2}+1}[x]^{1}_{1/2}$ $=\frac{1}{2y^{2}+2}$ then $\int^{2}_{0}\int^{1}_{1/2}\frac{1}{y^{2}+1}dydx=\int^{2}_{0}\frac{1}{2y^{2}+2}dx$ intergrating $[\frac{1}{2}\ln|2y^{2}+2|]^{2}_{0}$ $=\frac{1}{2}(\ln10-\ln2)$ $=\frac{1}{2}\ln5$