Evaluate a problem

Question

              ²

Evaluate ∫ |x-1|dx

             °

 

0

1
2
None of these

Answer 1

Recall that lAl = A if A≥0, but lAl = -A, if A<0.  

 

So, lx-1l = -(x-1) when x≤1, and lx-1l = x-1 when x≥1.

 

∫_{(from 0 to 2)}lx-1ldx = ∫_{(from 0 to 1)}lx-1ldx + ∫_{(from 1 to 2)}lx-1ldx

 

                           =∫_{(from 0 to 1)} (1-x)dx + ∫_{(from 1 to 2)}(x-1)dx

 

                           = (x-(1/2)x²)_{(from 0 to 1)} + ((1/2)x² - x)_{(from 1 to 2)}

 

                           = [1/2 - 0] + [(0 - (-1/2)] = 1

 

  

Answer 2

Make a sketch of the graph, and you will see that the area of two triangles is what this definite integral represents.  The answer is 1.

 

You don't need calculus for this one.