Intergration

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

Compute the definite integral: integral_(-2)^3 (x^3+x^2-6 x) dx Integrate the sum term by term and factor out constants: = integral_(-2)^3 x^3 dx+ integral_(-2)^3 x^2 dx-6 integral_(-2)^3 x dx Apply the fundamental theorem of calculus. The antiderivative of x^3 is x^4/4: = x^4/4|_(-2)^3+ integral_(-2)^3 x^2 dx-6 integral_(-2)^3 x dx Evaluate the antiderivative at the limits and subtract. x^4/4|_(-2)^3 = 3^4/4-(-2)^4/4 = 65/4: = 65/4+ integral_(-2)^3 x^2 dx-6 integral_(-2)^3 x dx Apply the fundamental theorem of calculus. The antiderivative of x^2 is x^3/3: = 65/4+x^3/3|_(-2)^3-6 integral_(-2)^3 x dx Evaluate the antiderivative at the limits and subtract. x^3/3|_(-2)^3 = 3^3/3-((-2)^3/3) = 35/3: = 335/12-6 integral_(-2)^3 x dx Apply the fundamental theorem of calculus. The antiderivative of x is x^2/2: = 335/12+(-3 x^2)|_(-2)^3 Evaluate the antiderivative at the limits and subtract. (-3 x^2)|_(-2)^3 = (-3 3^2)-(-3 (-2)^2) = -15: Answer: | | = 155/12

I have found that the answer is 21 and 1/12. But I can't figure out how to get there.