a. Explain why 0 ≤ x^2 tan^−1 x ≤ (πx^2)/ 4 for all 0 ≤ x ≤ 1. b. Use the properties of the integrals to show that the value of the integral ∫ x^2 tan^−1 x dx lies on the interval [0, π/12].

Question

In part b, ∫ x^2 tan^−1 x dx has bounds of 1(upper bound) and 0 (lower bound)

Answer 1

I'll use arctan(x) in place of tan^-1(x) here, but know they mean the same thing.

Part a.)

The absolute extrema of arctan(x) on the closed interval [0, 1] are: 0 as the minimum value and pi/4 as the maximum value.

To check this, you can find any local extrema on the open interval (0, 1) and compare all values along with the values of the function at the end points x=0 and x=1 to see what the least and greatest values are. (Note, this works only because arctan(x) is continuous on the interval). Another way you can show that 0 is the minimum value and pi/4 is the maximum value is by showing arctan(x) is strictly increasing on the interval [0, 1] and so the minimum must occur at x=0 (arctan(0) = 0) and the maximum must occur at x=1 (arctan(1) = pi/4).

Once you've shown that

0 ≤ arctan(x) ≤ pi/4 on the interval [0, 1],

you can multiply all sides of the inequality by x^2, achieving your desired result.

Part b.)

The integral property that is useful to us here: if f(x) ≤ g(x) on some interval [a, b] then

∫f(x) dx ≤ ∫g(x) dx

Note that the lower bound of both integrals is a and the upper bound of both is b, I just can't do that with this text editor.

Using our result from Part a.):

∫ 0 dx ≤ ∫ (x^2)arctan(x) dx ≤ ∫ (x^2)pi/4 dx

Integral of 0 from 0 to 1 is 0. By power rule the integral of (x^2)pi/4 from 0 to 1 is pi/12. Thus our inequality becomes the desired result:

0 ≤ ∫ (x^2)arctan(x) dx ≤ pi/12