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

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ggrree
  • ggrree
that's correct, chlorophyll! Could you please show your work? I am having trouble getting there.
nikvist
  • nikvist
check my work again. I think my work is correct
anonymous
  • anonymous
Just exactly as the nikvist's progress!
anonymous
  • anonymous
Take derivative then plug x = 3 in!
anonymous
  • anonymous
In mean Integral!
ggrree
  • 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.
ggrree
  • ggrree
I computed and got a negative value, which doesn't make sense :(
anonymous
  • anonymous
since when you sketch the graph, y = 3x is above y = 3x^2 - 6x
anonymous
  • anonymous
That's why you subtract: 3x - ( 3x^2 -6x)
ggrree
  • ggrree
|dw:1332618620433:dw|
anonymous
  • anonymous
Yep, then we calculate from the intersection part!
anonymous
  • anonymous
Make sense?
anonymous
  • anonymous
Next time, pls don't be hurry in make your comment that the solver is incorrect!
ggrree
  • 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: \[region = \int\limits_{0}^{3} 3x dx - \int\limits_{2}^{3} 3x^2 +3xdx - \int\limits_{0}^{2} 3x^2-3xdx\]
anonymous
  • anonymous
We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!
anonymous
  • anonymous
Since the line is above the curve in the interval [ 0, 3]
anonymous
  • anonymous
That's why you subtract: 3x - ( 3x^2 -6x) = -3x^2 + 9x
anonymous
  • anonymous
Are you with me?
ggrree
  • 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?
anonymous
  • anonymous
In case you have the straight line cut the third degree one, it's possible to have more than 1 limited portion!
anonymous
  • anonymous
In here, since quadratic just curve 1 time, so we're luckily have only 1 limited portion to calculate :)
anonymous
  • anonymous
I'm trying to explain as person somewhat gain more practice than the newbie!
anonymous
  • anonymous
This topic isn't as tough as you thought at all!
ggrree
  • ggrree
ok, so the formula I used is correct? (for things that only intersect twice to make one area)
anonymous
  • anonymous
Technically the formula is applied for all cases! It just different at how many closed area to be calculated!

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