Answer
Upper bound =0.9 and lower bound : 0.65
Work Step by Step
When the function $f(x)$ has minimum and maximum on $[a,b]$ then, we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$
Given: $f(x)=\dfrac{1}{1+x^2}$
on the interval $[0,0.5]$, max f=1 and min f=0.8
Now, we have $0.4 \leq \int_0^{0.5} \dfrac{1}{1+x^2} dx \leq 0.5$ ..(1)
on the interval $[0.5,1]$, max f=0.8 and min f=0.5
Now, we have $0.25 \leq \int_0^{0.5} \dfrac{1}{1+x^2} dx \leq 0.4$...(2)
After adding the equations (1) and (2), we have $0.65 \leq \int_0^{1} \dfrac{1}{1+x^2} dx \leq 0.9$
Thus, Upper bound =0.9 and Lower bound : 0.65
