Answer
$\frac{1}{4}+\frac{2}{5}\leq\int{0.5}{0}\frac{1}{1+(x)^2}dx+\int^{1}_{0.5}\frac{1}{1+(x)^2}dx\leq\frac{1}{2}+\frac{2}{5}=\frac{13}{20}\leq\int_{0}^{1}\frac{1}{1+(x)^2}dx\leq\frac{9}{10}$
Work Step by Step
--On[0,0,5],
-max f=$\frac{1}{1+(0)^2}=1$,
-min f=$\frac{1}{1+(0.5)^2}=0.8$,
therefore (0.5-0) min $f\leq\int^{0.5}_{0}f(x)dx\leq(0.5-0) $
-max f=$\frac{2}{5}\leq\int^{0.5}_{0}$ $\frac{1}{1+(x)^2}dx\leq\frac{1}{2}$
--on [0,5,1]
-max f=$\frac{1}{1+(0.5)^2}=0.8$
and
- min f=$\frac{1}{1+(1)^2}=0.5$
--therefore (1-0.5)
- min $\leq\int^{1}_{0.5}$ $\frac{1}{1+(x)^2}dx\leq(1-0.5)$
-max=$\frac{1}{4}\leq\int^{1}_{0.5}$ $\frac{1}{1+(x)^2}\leq\frac{2}{5}$
then
$\frac{1}{4}+\frac{2}{5}\leq\int{0.5}{0}\frac{1}{1+(x)^2}dx+\int^{1}_{0.5}\frac{1}{1+(x)^2}dx\leq\frac{1}{2}+\frac{2}{5}=\frac{13}{20}\leq\int_{0}^{1}\frac{1}{1+(x)^2}dx\leq\frac{9}{10}$
