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 m(b − a) ≤ b f(x) dx a ≤ M(b − a). Use this property to estimate the value of the integral. 1 0 x3 dx

Answer 1

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

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

Now we have to differentiate it

$f(x)' = \frac{(1 + x^2) \times 0 - 3\times 2x }{(1 + x^2)^2} \\\\ = \frac{-6x}{(1 + x^2)^2}$

Now the points i.e x = 0 and x =1 could be tested

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

= 3

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

= 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