Answer
upper bound=1
and
lower bound =$\frac{1}{2}$
Work Step by Step
$f(x)=\frac{1}{1-x^2}$ is decreasing on[0,1]
-maximum value that f occurs at 0 =>max f=f(01
-minimum value of f occurs at 1 =>min f=f(1)=$\frac{1}{1+1}=\frac{1}{2}$
-therefore(1-0) min f $\leq\int^{1}_{0}\frac{1}{1-x^2}dx\leq(1-0)$
-max f $\frac{1}{2}\leq\int^{1}_{0}\frac{1}{1+x^2}dx\leq1$
-that is an upper bound=1
and
-a lower bound =$\frac{1}{2}$
