Question

How do you find `int_-3^2 1/2(x+3)(x-2)?`

Answer 1

`int_(-3)^2 1/2(x+3)(x-2)dx=-125/12`

Explanation:

We will use the following:

• `int_a^b(f+g) = int_a^bf + int_a^bg` for integrable functions `f` and `g`
• `int_a^bcf = cint_a^bf` for an integrable function `f` and constant `c`
• `int_a^bx^ndx = (b^(n+1)-a^(n+1))/(n+1)` for `n!=-1`

With those, we have:

`int_(-3)^2 1/2(x+3)(x-2)dx = 1/2int_(-3)^2(x+3)(x-2)dx`

`=1/2int_(-3)^2(x^2+x-6)dx`

`=1/2(int_(-3)^2x^2dx + int_(-3)^2x^1dx - 6int_(-3)^2x^0dx)`

`=1/2((2^3-(-3)^3)/3 + (2^2-(-3)^2)/2-(6(2-(-3)))/1)`

`=1/2(35/3 - 5/2 - 30)`

`=-125/12`