Question

What is `f(x) = int e^x-x dx` if `f(-1) = 1`?

Answer 1

`f(x)=e^(x)-x^2/2+3/2-e^{-1}`

Explanation:

The general antiderivative is

`\int (e^{x}-x)\ dx=e^{x}-x^{2}/2+C`

Therefore, `f(x)=e^{x}-x^{2}/2+C` for some constant `C`. Since `f(-1)=1`, we get `1=e^{-1}-(-1)^{2}/2+C=e^{-1}-1/2+C`, which implies that `C=3/2-e^{-1}`.

Hence, `f(x)=e^(x)-x^2/2+3/2-e^{-1}`.