Question

How do you evaluate an integral with an absolute value? Question below.

Answer 1

`int_0^1 abs(2x^2+x-1)*dx=3/4`

Explanation:

Once I found roots of integrand,

`2x^2+x-1=0`

`(2x-1)*(x+1)=0`

Hence `x_1=-1` and `x_2=1/2`

Integrand was negative when `-1<=x<=1/2` and positive in other situations.

Consequently,

`int_0^1 abs(2x^2+x-1)*dx`

=`int_0^(1/2) (-2x^2-x+1)*dx`+`int_(1/2)^1 (2x^2+x-1)*dx`

`A=int_0^(1/2) (-2x^2-x+1)*dx`

=`[-2/3x^3-1/2x^2+x]_0^(1/2)`

=`(-2/3)*1/8-(1/2)^3+1/2`

=`1/2-1/8-1/12`

=`7/24`

`B=int_(1/2)^1 (2x^2+x-1)*dx`

=`[2/3x^3+1/2x^2-x]_(1/2)^1`

=`2/3*(1-1/8)+1/2*(1-1/4)-(1-1/2)`

=`7/12+3/8-1/2`

=`11/24`

Thus,

`int_0^1 abs(2x^2+x-1)*dx`

=`A+B`

=`7/24+11/24`

=`3/4`