Evaluate: ∫52(|x-2|+|x-3|+|x-5|)dx

Let I = ∫52(|x-2|+|x-3|+|x-5|)dx = 1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5| (using indefinite integral form).
By the fundamental theorem of calculus ∫baF(x)dx = f(b) - f(a), so just evaluate integral endpoints and that's the answer (1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=5) - (1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=2) = 132
(1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=5) - (1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=2) = 11.5
I = ∫52(|x-2|+|x-3|+|x-5|)dx = (1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=5) - (1/2(x-2)|x-2| + 1/2(x-3)|x-3| + 1/2(x-5)|x-5|)|(x=3) = 232
I = ∫52(|x-2|+|x-3|+|x-5|)dx = 232 = 11.5