Find the are of the region enclosed between the graphs y = 3x and y =3x^2 - 6x

I know the answer, and I think I know how to solve it, but I'm having trouble computing the actual answer. Help would be appreciated.

Question

Find the are of the region enclosed between the graphs y = 3x and y =3x^2 - 6x

I know the answer, and I think I know how to solve it, but I'm having trouble computing the actual answer. Help would be appreciated.

anonymous · Answer

= 13.5

nikvist · Answer

$$y=3x\quad,\quad y=3x^2-6x$$$$3x^2-6x=3x$$$$3x^2-9x=0\quad;\quad x_1=0\quad,\quad x_2=3$$$$S=\int\limits_0^3(9x-3x^2)dx$$

ggrree · Answer

is that all there is to it, nikvist ?
A portion of the region lies below the x axis, so I think that your answer is incorrect.

ggrree · Answer

that's correct, chlorophyll! 
Could you please show your work? I am having trouble getting there.

nikvist · Answer

check my work again. I think my work is correct

anonymous · Answer

Just exactly as the nikvist's progress!

anonymous · Answer

Take derivative then plug x = 3 in!

anonymous · Answer

In mean Integral!

ggrree · Answer

I am not understanding how he got $$\int\limits_{0}^{3} 9x - 3x^2 dx$$.
Why are the signs reversed here? shouldn't it at the very least be $$\int\limits_{0}^{3} 3x^2 - 9x dx$$ ?
since that is what he wrote the line before.

ggrree · Answer

I computed and got a negative value, which doesn't make sense :(

anonymous · Answer

since when you sketch the graph, y = 3x is above y = 3x^2 - 6x

anonymous · Answer

That's why you subtract: 3x - ( 3x^2 -6x)

ggrree · Answer

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