Integrate [ x^3+x^2-6x] from -2 to 3
We need to split this integral up into three parts
x^4/4 + x^3/3 - 3x^2 from -2 to 0
- [(-2)^4/4 + (-2)^3/3 - 3(-2)^2] ] =
-4 + 8/3 + 12 = -4 + 8/3 + 12 = 8 + 8/3 = 10 2/3 = 32/3
Integrate [ x^3+x^2-6x] from 0 to 2
x^4/4 + x^3/3 - 3x^2 from 0 to 2
(2)^4/4 + (2)^3/3 - 3(2)^2 =
4 + 8/3 - 12 =
20/3 - 36/3 = -16/3
Integrate [ x^3+x^2-6x] from 2 to 3 =
x^4/4 + x^3/3 - 3x^2 from 2 to 3
3^4/4 + 3^3/3 - 3(3)^2 - [-16/3] =
81/4 + 27/3 - 27 + 16/3 =
81/4 - 18 + 16/3
243/12 - 216/12 + 64/12 = 91/12
So....the whole area =
32/3 - 16/3 + 91/12 =
16/3 + 91/12 =
64/12 + 91/12 =
155/12 =
12 + 11/12
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