## anonymous 4 years ago 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.

1. anonymous

= 13.5

2. anonymous

$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$

3. anonymous

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.

4. anonymous

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

5. anonymous

check my work again. I think my work is correct

6. anonymous

Just exactly as the nikvist's progress!

7. anonymous

Take derivative then plug x = 3 in!

8. anonymous

In mean Integral!

9. anonymous

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.

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

|dw:1332618620433:dw|

14. anonymous

Yep, then we calculate from the intersection part!

15. anonymous

Make sense?

16. anonymous

Next time, pls don't be hurry in make your comment that the solver is incorrect!

17. anonymous

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: $region = \int\limits_{0}^{3} 3x dx - \int\limits_{2}^{3} 3x^2 +3xdx - \int\limits_{0}^{2} 3x^2-3xdx$

18. anonymous

We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!

19. anonymous

Since the line is above the curve in the interval [ 0, 3]

20. anonymous

That's why you subtract: 3x - ( 3x^2 -6x) = -3x^2 + 9x

21. anonymous

Are you with me?

22. anonymous

"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?

23. anonymous

In case you have the straight line cut the third degree one, it's possible to have more than 1 limited portion!

24. anonymous

In here, since quadratic just curve 1 time, so we're luckily have only 1 limited portion to calculate :)

25. anonymous

I'm trying to explain as person somewhat gain more practice than the newbie!

26. anonymous

This topic isn't as tough as you thought at all!

27. anonymous

ok, so the formula I used is correct? (for things that only intersect twice to make one area)

28. anonymous

Technically the formula is applied for all cases! It just different at how many closed area to be calculated!