Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 56

You need to check if int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx , using mean value theorem, such that:
int_a^b f(x)dx = (b-a)f(c), where c in (a,b)
int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx => int_0^1 sqrt(1+x^2)dx - int_0^1sqrt(1+x)dx <= 0
int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx <= 0
According to mean value theorem yields:
int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx = (1-0)f(c), c in (0,1)
You need to verify the monotony of the function f(x) = sqrt(1+x^2)dx - sqrt(1+x), such that:
f'(x) = x/(sqrt(1+x^2)) - 1/(2sqrt(1+x))
f'(x) = (2x*sqrt(1+x)-sqrt(1+x^2)) /(2sqrt(1+x^2)(1+x))
f'(x) = 0 => 2x*sqrt(1+x)-sqrt(1+x^2) = 0 => 2x*sqrt(1+x)=sqrt(1+x^2)
Raising to square:

4x^2(1+x) = 1+x^2
4x^2 + 4x^3 - 1 - x^2 = 0
4x^3 + 3x^2 - 1 >0 for x in (0,1) => f(x) increases on (0,1)
For c in (0,1) => 0 f(0)f(0) = sqrt1 - sqrt1 = 0
f(1) = sqrt(1+1) - sqrt(1+1)=0
Since f(c) < f(1) = 0 => f(c) < 0 . Since f(c) = int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx , then int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx< 0.
Hence, checking if the inequality holds, using mean value theorem, yields int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx is verified.