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anonymous
 5 years ago
area between two graphs. y=2 and y=x^22
anonymous
 5 years ago
area between two graphs. y=2 and y=x^22

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know how to integrate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. So to find the area under the function you just integrate it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set the two equations equal to find the bounds

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since the area under the top function is the area between them and the area under the bottom function. The area between them would be Area under top function  Area under bottom function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So find the x values for which the two functions are the same, then integrate each function between those bounds, and subtract the resulting areas. Sometimes it's easiest to combine the two integrals and simplify before evaluating the integral, but sometimes it's not.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be 10.66 repeating?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, since your parabola is under the y=0 line for a bit this method won't quite work. You'll need to break up the integrals. You'll have \[Area = \int\limits_{2}^{\sqrt{2}}[2  (x^22)]dx  \int\limits_{\sqrt{2}}^{\sqrt{2}}(x^22)dx + \int\limits_{\sqrt{2}}^{2}[2  (x^22)]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats nonsence... just integrate \[\int\limits\limits_{2}^{2}2x ^{2}+2 dx=\int\limits\limits_{2}^{2}4x ^{2} dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That doesn't work. Because part of that integral will be negative area.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, since you are adding 2 to the parabola in fact. it never crosses xaxis

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for practice, let me see what I can mess up :) I agree that the interval is [2,2], but since we have mirror images across the yaxis, we can just do [0,2] and double it. y = 2 is just a box that is 2(2 by 2) = 8 (S)x^22 dx > x^3/3 2x 0,2 = 2[(8/3) 6] = 16/3 36/3 = 20/3 8 20/3 = 24/3 + 20/3 = 44/3 I would say, and this is my guess.. that the area between the 2 curves is 44/3. Now just let me know what I did wrong :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0everything :) \[x^{3}/3  2x \] from 0 to 2 is 8/34=8/312/3=4/3 *2 => 8/3 8(8/3)=32/3=10+2/3=10.66666.... :D
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