1-1-22-2-4-6-8-10xy

The curve y = x3, showing the portion under the curve from x = −2 to x = 1.

We can see from the graph that the portion between `x = -2` and `x = 0` is below the x-axis, so we need to take the absolute value for that portion.

`text[Area]= |int_-2^0x^3 dx|+int_0^1x^3 dx`

`=|[x^4/4]_-2^0|+[x^4/4]_0^1`

`=|(0-16/4)|+(1/4-0)`

`=4+1/4`

`=4.25\ text[units]^2`

Easy to understand math videos:
MathTutorDVD.com