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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.

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= 13.5
\[y=3x\quad,\quad y=3x^2-6x\]\[3x^2-6x=3x\]\[3x^2-9x=0\quad;\quad x_1=0\quad,\quad x_2=3\]\[S=\int\limits_0^3(9x-3x^2)dx\]
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.

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Other answers:

that's correct, chlorophyll! Could you please show your work? I am having trouble getting there.
check my work again. I think my work is correct
Just exactly as the nikvist's progress!
Take derivative then plug x = 3 in!
In mean Integral!
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.
I computed and got a negative value, which doesn't make sense :(
since when you sketch the graph, y = 3x is above y = 3x^2 - 6x
That's why you subtract: 3x - ( 3x^2 -6x)
Yep, then we calculate from the intersection part!
Make sense?
Next time, pls don't be hurry in make your comment that the solver is incorrect!
of course I don't mean to be disrespectful at all( I am very grateful that you are helping me :) ) but there are things that I don't understand. When I did this problem I looked at the graph and saw that a part was under the graph, so I did: \[region = \int\limits_{0}^{3} 3x dx - \int\limits_{2}^{3} 3x^2 +3xdx - \int\limits_{0}^{2} 3x^2-3xdx\]
We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!
Since the line is above the curve in the interval [ 0, 3]
That's why you subtract: 3x - ( 3x^2 -6x) = -3x^2 + 9x
Are you with me?
"We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!" This I don't understand... I understand that in general, if you want to find the area of a region enclosed by two curves f(x) and g(x) which intersect at points A and B, and f(x) > g(x) you must do \[\int\limits_{A}^{B} f(x) - g(x) dx\] but here a part of the first graph is below the x axis, so wouldn't the integral there be negative?
In case you have the straight line cut the third degree one, it's possible to have more than 1 limited portion!
In here, since quadratic just curve 1 time, so we're luckily have only 1 limited portion to calculate :)
I'm trying to explain as person somewhat gain more practice than the newbie!
This topic isn't as tough as you thought at all!
ok, so the formula I used is correct? (for things that only intersect twice to make one area)
Technically the formula is applied for all cases! It just different at how many closed area to be calculated!

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