Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 55

You need to use the mean value thorem to verify the given inequality, such that:
int_a^b f(x)dx = (b-a)*f(c), c in (a,b)
Replacing x^2 - 4x + 4 for f(x) and 0 for a, 4 for b, yields:
int_0^4 (x^2 - 4x + 4) dx= (4-0)f(c) = 4f(c)
You need to prove that int_0^4 (x^2 - 4x + 4) dx >= 0, hence, according to mean value theorem, 4f(c)>=0.
You need to check the monotony of the function f(x), hence, you need to find the derivative using the chain rule and to verify if it is positive or negative on interval (0,4), such that:
f(x) = (x^2 - 4x + 4) => f(x) = (x-2)^2 => f'(x) = 2(x-2)*(x-2)'
f'(x) = 2(x-2)
You need to notice that f(x) decreases on (0,2) and it increases on (2,4).
Hence, for c in (0,2) yields:
0 f(0)>f(c)>f(2)
You need to evalute f(0) and f(2), such that:
f(0) = (0-2)^2 = 4
f(2) = (2-2)^2 = 0
Hence, replacing the found values in inequality f(0)>f(c)>f(2) , yields:
4> f(c)>0
You need to obtain 4f(c), hence, multiplying by positive 4 both sides yields:
16> 4f(c)>0
But 4f(c) = int_0^4 (x^2 - 4x + 4) dx and 4f(c)>0, hence int_0^4 (x^2 - 4x + 4) dx >0.
Hence, using mean value theorem, yields that the inequality int_0^4 (x^2 - 4x + 4) dx >0 holds. Notice that it is no need to evaluate the integral to prove the inequality.