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anonymous
 4 years ago
Find the postive value of c such that the area of the the region enclosed by the parabolas
y = x^2 c^2 and y = c^2 x^2 is 72.
anonymous
 4 years ago
Find the postive value of c such that the area of the the region enclosed by the parabolas y = x^2 c^2 and y = c^2 x^2 is 72.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I thought I had to show c^2=x^2 so x=+c so I evaluate from c to c. But the equation is throwing me off if I was even right.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah that's right so far, let me make sure I have the answer though...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok, we got the top graph is c^2x^2, and the bottom is x^2c^2, so our integral is\[\int_{c}^{c}c^2x^2(x^2c^2)dx=\int_{c}^{c}2c^22x^2dx\]the integrand is even so we can simplify this a bit\[=2\int_{0}^{c}2c^22x^2=2(2c^2x\frac23x^3)_{0}^{c}=2(2c^3\frac23c^3)=72\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh I was doing it the other way. I put f(x) to be (x^2c^2) and g(x) to be the other one and was getting a werid answer.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, it's a really good idea to sketch these thing out when you're not 100% on them. many people neglect that, and it costs them. welcome!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Speaking of sketching, can you check what I may have done wrong here. It says y=x^2 and y=82x find area between xaxis and 2 curves. So I did \[\int\limits_{4}^{2}x^2(82x)\] it gives me a negative area?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah it sure does... think about this: what is x^2 when x=0 ? what is 82x when x=0 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So wait did I do it wrong? I thought I should find where they intercept and then take the intergral. Iver never done one like this with the "between xaxis" part so not sure.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yes, but just answer my question let f(x)=x^2 let g(x)=82x what is f(0) ? what is g(0) ? picking a point in your interval of integration to check will tell you which graph should be on top and which should be on the bottom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well x^2 would be 0 and 82(0) would be 4

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so 82x should be on the top, yes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh so it should be (82x)(x2) instead?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yep 82x>x^2 on the interval you are talking about http://www.wolframalpha.com/input/?i=plot+x%5E2%2C82x

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1...hence the negative answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay let me try this and see if it is better

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dang I wish I knew all this like you ha would make this extensive homework easier

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Just practice is all. What I study is still hard for me 'cause it's new. By the time you learn it it's on to the next thing :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is only one more question I had i doubt i can explain it here though. Let me try quick.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When you are finding the area of 2 lines with 3 shaded regions, how do you find the middle one when points arent given? the lines are y=x/3 and y=x^3/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can get the first region because i just evaluate from 2 to 0. The rest never seen before since some is above the x/3 line and some before

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I only see two shaded regions between these graphs: between 1,0 and 0,1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Our drawing is from 2 and 0, 0 to 3 ish? but some is above the x/3 line and some below

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac x3=\frac{x^3}3\to x=x^3\to x(1x^2)=0\to x=\left\{ 1,0,1 \right\}\]are the intervals here's a visual http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328632999874:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not the best drawing but it wants the regions

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1click the link for a better image of the graph

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1does it tell you it wants the region between the graphs, or does it give you a specific interval to integrate?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The shaded region I drew. I think thats the region between graphs?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1but the last region you drew on the right is unbounded: it goes to infinity again click the link I posted, there are only 2 regions enclosed between the graphs: http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well i think its bounded between the line and x=3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so he wants the area from the line at x=3 so it now has bounds i guess

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1then your intervals are\[[1,0][0,1][1,3]\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so you need to make sure that in each interval you have the proper graph on top, and proper graph on bottom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are u sure its not 2,0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my book has the points (2,2/3) as one point and (3,6) as another

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1if you want the region between the graphs then\[\frac{2}{3}\neq\frac{(2)^3}3\]so no

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1at least that's not a point of intersection

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1if you are told to integrate from x=2 to x=3 then so be it, but those are not intersection points...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i feel stupid sorry. the line is x^3/3 x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0should make more sense

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{x^3}3x=\frac x3\to\frac{x^3}3=\frac{4x}{3}\to x(x^24)=0\]\[x=\left\{ 2,0,2 \right\}\]so it's still not the right interval :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well the 3rd shaded region the lines dont intercept

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its just from the point of (3,6) straight down to (3,1)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok, so we have from\[[2,0][0,2][2,3]\]look at the graph http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3x on the first interval x^3/3x is on top on the second, x/3 is on top on the third, x^3/3 is on top again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to go eat lunch, thanks for the help though. I will run it past my professor and maybe will get help before its due. Thanks

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1me too, got class you should be able to set up the integral now though, good luck!

myname
 4 years ago
Best ResponseYou've already chosen the best response.0Is c a constant or a variable?
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