Question

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then

Answer 1

Answer:

smaller value is 1.2

larger value is 6

Step-by-step explanation:

$f(x)=\frac{3}{1+x^{2} }$

when x=a=0, one has

$f(0)=\frac{3}{1+0^{2} } \\f(0)=3$

now, when x=b=2, one has

$f(2)=\frac{3}{1+2^{2} } \\f(2)=\frac{3}{5}$

Therefore, the absolute minumun is

$m=\frac{3}{5}$

and the absolute maximun is

$M=3$

The approximation to the integral is

$\frac{3}{5}(2-0)\leq \int\limits^2_0 {f(x)} \, dx \leq 3(2-0)$

hence

$\frac{3}{5}(2)\leq \int\limits^2_0 {f(x)} \, dx \leq 3(2)\\\frac{6}{5} \leq \int\limits^2_0 {f(x)} \, dx \leq 6\\1.2 \leq \int\limits^2_0 {f(x)} \, dx \leq 6$