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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.
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.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=3x\quad,\quad y=3x^26x\]\[3x^26x=3x\]\[3x^29x=0\quad;\quad x_1=0\quad,\quad x_2=3\]\[S=\int\limits_0^3(9x3x^2)dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's correct, chlorophyll! Could you please show your work? I am having trouble getting there.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0check my work again. I think my work is correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just exactly as the nikvist's progress!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take derivative then plug x = 3 in!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I computed and got a negative value, which doesn't make sense :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since when you sketch the graph, y = 3x is above y = 3x^2  6x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's why you subtract: 3x  ( 3x^2 6x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1332618620433:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, then we calculate from the intersection part!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Next time, pls don't be hurry in make your comment that the solver is incorrect!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0of 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: \[region = \int\limits_{0}^{3} 3x dx  \int\limits_{2}^{3} 3x^2 +3xdx  \int\limits_{0}^{2} 3x^23xdx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since the line is above the curve in the interval [ 0, 3]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's why you subtract: 3x  ( 3x^2 6x) = 3x^2 + 9x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"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
 4 years ago
Best ResponseYou've already chosen the best response.0In case you have the straight line cut the third degree one, it's possible to have more than 1 limited portion!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In here, since quadratic just curve 1 time, so we're luckily have only 1 limited portion to calculate :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm trying to explain as person somewhat gain more practice than the newbie!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This topic isn't as tough as you thought at all!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, so the formula I used is correct? (for things that only intersect twice to make one area)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Technically the formula is applied for all cases! It just different at how many closed area to be calculated!
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