How do you integrate with absolute value

From 0 to 2. Integral of the absolute value of (x+3)^(1/2)-2x dx

My instinct says (2/3)(x+3)^(3/2) - 2x^2 /2. But I think that would be wrong?

we should note that the function is 0 at x = 1 , > 0 on [0,1] & < 0 on [1,2]

thus int over x in [0,1] of { √(x+3) - 2x } dx + int over x in [1,2] of { - (√(x+3) - 2x ) } dx

{ 38/3 - 2 √3 - 2 √5 }

You just integrated normally. The absolute value mean you just switch all the negatives into positives. So...

Given:
Integral of the absolute value of (x+3)^(1/2)-2x dx

Take the absolute value of everything gives us:
((x+3)^(1/2)) + 2x From 0 to 2.

Now, It's incredibly easy to integrate. Break up the integral into two integrals resulting in:
Integral 1: (root(x+3) dx
Integral 2: 2x dx
both with limits from 0 to 2

Anti-derivative of integral 1: 2(x+3)^(3/2)
Anti-derivative of integral 2: x^2

Combine both to get:

(2((x+3)^(3/2)) + (x^2) from 0 to 2

F(2) - F(0) = 10(root(5)) - 6(root(3)) + 4