Question

How do you find `\int ( 13- x ) d x`?

Answer 1

`-1/2x^2+13x+C_1-C_2`, where `C_1` and `C_2` are constants.

Explanation:

Given: `int(13-x)dx`

We use the difference rule, which states that

`int(a-b)dx=inta \ dx-intb \ dx`

`=int13 \ dx-intx \ dx`

`=(13x+C_1)-(1/2x^2+C_2)`

`=13x-1/2x^2+C_1-C_2`