Question

How do you find `\int _ { 1} ^ { 3} ( x ^ { 2} - 1) d x`?

Answer 1

`20/3`

Explanation:

Since `F(x)=int(x^2-1)=x^3/3-x+c`

you would get

`int_a^b(x^2-1)=F(3)-F(1)=(3^3/3-3)-(1^3/3-1)`

`=9-3-1/3+1`

`=20/3`