## ggrree Group Title 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. 2 years ago 2 years ago

1. Chlorophyll

= 13.5

2. nikvist

$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. ggrree

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. ggrree

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

5. nikvist

check my work again. I think my work is correct

6. Chlorophyll

Just exactly as the nikvist's progress!

7. Chlorophyll

Take derivative then plug x = 3 in!

8. Chlorophyll

In mean Integral!

9. ggrree

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. ggrree

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

11. Chlorophyll

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

12. Chlorophyll

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

13. ggrree

|dw:1332618620433:dw|

14. Chlorophyll

Yep, then we calculate from the intersection part!

15. Chlorophyll

Make sense?

16. Chlorophyll

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

17. ggrree

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. Chlorophyll

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

19. Chlorophyll

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

20. Chlorophyll

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

21. Chlorophyll

Are you with me?

22. ggrree

"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. Chlorophyll

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

24. Chlorophyll

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

25. Chlorophyll

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

26. Chlorophyll

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

27. ggrree

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

28. Chlorophyll

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