Answer:
m = 1.5
M = 3
Step-by-step explanation:
According to the situation, the solution is as follows
Here we need to determine the maximum and minimum of
[tex]f(x) = \frac{3}{1 + x^2}\ on\ 0,1[/tex]
Now we have to differentiate it
[tex]f(x)' = \frac{(1 + x^2) \times 0 - 3\times 2x }{(1 + x^2)^2} \\\\ = \frac{-6x}{(1 + x^2)^2}[/tex]
Now the points i.e x = 0 and x =1 could be tested
[tex]f(0) = \frac{3}{1 + 0}[/tex]
= 3
[tex]f(1) = \frac{3}{1 + 1}[/tex]
= 1.5
So,
m = 1.5
M = 3
By applying the differentiation we could easily estimate the value of the integral and the same is to be shown above