## University Calculus: Early Transcendentals (3rd Edition)

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