Question

How do you evaluate the definite integral `int (x^2-3x+8)dx` from [-2,5]?

Answer 1

`413/6`

Explanation:

Integrate each term using the `color(blue)"power rule for integration"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(int(ax^n)=a/(n+1)x^(n+1))color(white)(2/2)|)))`

`rArrint_(-2)^5(x^2-3x+8)dx`

`=[1/3x^3-3/2x^2+8x]_(-2)^5`

Evaluate at x = 5 and subtract the evaluation at x = - 2

`=(125/3-75/2+40)-(-8/3-6-16)`

`=265/6-(-74/3)=265/6+148/6=413/6`