Find the integral?

Question

Find the integral of abs(x) dx from -1 to 3.

 

abs means absolute value.

 

Answer: 5

 

Please show all your work.

Answer 1

|x| = x for x≥0, -x for x<0.  So break the integral into two parts.

 

 3                0              3                          0                    3     

 ∫ |x|dx =  ∫ -xdx + ∫ xdx = -(1/2)x² |  + (1/2)x² |  =  1/2 + 9/2 = 10/2 = 5
-1               -1             0                          -1                   0

                      

Answer 2

 

 

Integral ( -1 to 3, |x| dx )

Remember the definition of absolute value.

|x| = { x, if x >= 0

         { -x, if x < 0


Integral (-1 to 0, |x| dx) + Integral (0 to 3, |x|  dx )

You can change |x| into -x for the first integral, and |x| into x in the second integral.

Integral (-1 to 0, (-x) dx) + Integral (0 to 3, x dx )


 Integral (-1 to 0, x dx) + Integral (0 to 3, x dx )

Solve as normal.

[ (1/2)x^2) ] {evaluated from -1 to 0} + [ (1/2)x^2 ] {evaluated from 0 to 3}

 { (1/2)(0)^2 - (1/2)(-1)^2 } + { (1/2)(3)^2 - (1/2)(0)^2 }

Then solve this ....