Answer:
smaller value is 1.2
larger value is 6
Step-by-step explanation:
[tex]f(x)=\frac{3}{1+x^{2} }[/tex]
when x=a=0, one has
[tex]f(0)=\frac{3}{1+0^{2} } \\f(0)=3[/tex]
now, when x=b=2, one has
[tex]f(2)=\frac{3}{1+2^{2} } \\f(2)=\frac{3}{5}[/tex]
Therefore, the absolute minumun is
[tex]m=\frac{3}{5}[/tex]
and the absolute maximun is
[tex]M=3[/tex]
The approximation to the integral is
[tex]\frac{3}{5}(2-0)\leq \int\limits^2_0 {f(x)} \, dx \leq 3(2-0)[/tex]
hence
[tex]\frac{3}{5}(2)\leq \int\limits^2_0 {f(x)} \, dx \leq 3(2)\\\frac{6}{5} \leq \int\limits^2_0 {f(x)} \, dx \leq 6\\1.2 \leq \int\limits^2_0 {f(x)} \, dx \leq 6[/tex]
