Area of f: int [4,-2] (|3-|3x||)dx

[confused_07]

Given int [4,-2] (|3-|3x||)dx, fisrt sketch the graph y=f[x] on the given interval. Then find the integral of f using your knowledge of area formulas for rectangles, triangles, and circles.

I sketched the graph and came up with 3 triangles that connect using (-2,3), (-1,0), (0,3), (1,0), (4,9). (I don't know how to post pictures here). Using the area of a triangle (A=1/2bh), I got:

Triangle (-2,3)(-1,0)(-2,0)= 1.5
Triangle (-1,0)(0,3)(1,0)= 3
Triangle (1,0)(4,9)(4,0)= 13.5
Total area of all three = 18

For the integral:

=[|3x-|(3/2)x^2||] [4,-2]
=[|3(4)-|(3/2)(4^2)||] - [|3(-2)-|(3/2)(-2^2)||]
=(12-12)
=0

I am confused. What did I do wrong?

 

[galactus]

I have found it's sometimes best to use separate integrals to evaluate an absolute value.

Graph your function. You can see it's made up of lines. Calculate the areas of these regions separately. You will see it's 18.

Since there's symmetry:

\(\displaystyle \L\\3\int_{0}^{1}(-3x+3)dx+\int_{1}^{4}(3x-3)dx=\frac{36}{2}=18\)

 

[confused_07]

Thank you.....