absolute value integral

[hank]

Ok, the problem is this:

\(\displaystyle \int_0^2 {\left| {2x - 3} \right|dx}\)

Ok, so when x is positive, I should find the integral of 2x - 3.
When x is negative, I should find the integral of 3 - 2x.

However, since x is always positive, it seems to me I only need to find the integral of 2x - 3.

When I do that and plug in 2 and 0, I get -2.

However, the book says the answer is 5/2.

What am I missing?

 

[galactus]

\(\displaystyle \L\\\int_{0}^{\frac{3}{2}}{-}(2x-3)dx+\int_{\frac{3}{2}}^{2}2x-3dx\)

 

[hank]

Ok, so you set the 2x - 3 to zero and solve for x, and that's where the intervals are set?

 

[galactus]

Yes, that's correct. Graph, graph, graph. It'll help you see.