calculus - Calculation of definite integral

If someone could please help me with this question, I'm not sure where I've gone wrong.

I have the following definite integral

$$\int_0^9 {\sqrt {{({2x-6})^2}+{({2x-6})^2}}} \,dx \\$$

$$\int_0^9 { \sqrt {{2({2x-6})^2}}} \,dx \\$$

$$\int_0^9 { \sqrt {{2}}({2x-6})} \,dx \\$$

$$\sqrt {{2}} \int_0^9 { ({2x-6})} \,dx \\$$

So after integrating, I obtain the following: (I checked this with online integral calculators as well)

$$\ {\sqrt{2} } (x^2-6x)\, \\$$

So if I am evaluating this integral from 0 to 9, would my answer not be $$\ {27\sqrt{2} }\, \\$$

I don't know why the answer is

$$\ {45\sqrt{2} }\, \\$$

Anyone know where I went wrong?

Answer

Don't forget the absolute value.
$$\sqrt{2}\int_0^9|2x-6|dx=\sqrt{2}\left(\int_0^3(6-2x)dx+\int_3^9(2x-6)dx\right)$$
which is equal to $$\sqrt{2}\left(6x-x^2|_0^3+x^2-6x|_3^9\right)=45\sqrt{2}$$