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

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Answer 1 · 2018-03-17T14:57:50

$1/3$

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 · 2018-03-17T14:58:24

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

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

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

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

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

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