Question

What is `F(x) = int x-e^(-x) dx` if `F(0) = 2`?

Answer 1

`F(x)=x^2/2+e^(-x)+1`

Explanation:

`F(x)=intx-e^(-x)dx`
`F(x)=x^2/2+e^(-x)+C` where C is a constant number

Let's find C given `F(0)=2`

`F(0)=2`

`(0)^2/2+e^(-0)+C=2`

`0+1+C=2`
`C=2-1`
`C=1`

Therefore, `F(x)=x^2/2+e^(-x)+1`