Question

How do you evaluate the integral of `int x/6 dx` from 0 to 2?

Answer 1

`1/3`

Explanation:

Ok first the `1/6` is a constant so it can be pulled out of the integral and be re-written like so
`1/6int_0^2xdx`

The integral of `x` is `1/2x^2`

Now we multiply the `1/6` to the integral which is `1/2x^2`

Giving us
`1/12x^2`

Now we plug in the `x` values which were `2` and `0` and subtract them

`1/12(2)^2 - 1/12(0)^2`

Equals
`4/12-0/12`

`4/12`

Simplifying to
`1/3`

Answer 2

`int_0^2 x/6 dx = 1/3`

Explanation:

`int_0^2 x/6 dx = 1/6 int_0^2 xdx`

`int_0^2 x/6 dx = 1/6 [x^2/2]_0^2`

`int_0^2 x/6 dx = 1/6 (2^2/2-0)`

`int_0^2 x/6 dx = 1/3`