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anonymous
 5 years ago
find the area between the curves f(x)=(2x)^2 and g(x)=square root of x for the interval [0,3]
anonymous
 5 years ago
find the area between the curves f(x)=(2x)^2 and g(x)=square root of x for the interval [0,3]

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TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4we are going to need the point where these two graphs intersect, because they will cross each other, can you find that?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4dw:1327193397919:dwwe need the shaded area. Since area is always positive, we always need to subtract the lower area from the upper one, which meant we will need to split up our integral at the point where the graphs cross. To find that figure out when\[f(x)=g(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I do but I don't get the part the area for the interval [0,3]. Like we need to find the area of the two portions, one from 0 to 1 and one from 1 to 3 or what????

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4yes those are the right intervals

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use Riemann sums, Integration is overrated :P

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4so which interval will we want to use\[\int f(x)g(x)dx\]and which will we use\[\int g(x)f(x)dx\]?? @FFM I'm actually reviewing that now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Great, which one do you like most the right or the left one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hehe :D Can't we use there median? :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey wait wait wait for the interval [0,1] we use \[\int\limits_{0}^{1} f(x)g(x)\] and [1,3] we use g(x)f(x) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Turing is a awesome teacher! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@turing test.. your graph is not right... f(x)=(2x)^2 have minima at x=2, and also x=2 is the solution... after that it may not cut squareroot of x before 3

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4thx, only to good students @sam I really was only trying to illustrate the intersection, there is no scale, so it flies for our purposes

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4(1,1) is the intersection

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to find that area by calculator?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4no, I know how to find it by integrating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how? I can't find the area for [1,3] by calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YOu could use Mathematica :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Turning Test: thanks for your help. :) I'm trying to get the answer. FoolForMath: keke i know right. but it takes so long

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4\[\int_{1}^{3}g(x)f(x)dx=\int_{1}^{3}x^{1/2}(2x)^2dx=\int_{1}^{3}x^{1/2}(44x+x^2)dx\]got it from here?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4\[=\int_{1}^{3}x^{1/2}(44x+x^2)dx=\int_{1}^{3}x^{1/2}4+4xx^2dx\]...equals blah blah It's just algebra after integrating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I'm confused now ~.~

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4where did you get stuck? did you understand everything I typed so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you didn't type the one from [0,1), did you?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4no, I though you only were having trouble with the second Is that just because you did the other with a calculator? lol ...I'll type out what we have so far while you respond.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) hey I got the answer but it's probably wrong. cuz I got the answer from my teacher. my answer is 2.46 and his answer is 3.797.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4\[\int_{0}^{1}f(x)g(x)dx+\int_{1}^{3}g(x)f(x)dx\]\[=\int_{0}^{1}(2x)^2x^{1/2}dx+\int_{1}^{3}x^{1/2}(2x)^2dx\]\[\int_{0}^{1}24x+x^2x^{1/2}dx+\int_{1}^{3}x^{1/2}4+4xx^2dx\]...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4\[=2x2x^2+\frac{1}{3}x^3\frac{2}{3}x^{3/2}_{0}^{1}+\frac{2}{3}x^{3/2}4x+2x^2\frac{1}{3}x^3_{1}^{3}\]\[=(22+\frac 1 3\frac 2 3)+[\frac2 3(3\sqrt3)4(3)+2(9)][\frac2 3 4+2\frac1 3]\]...just simplifying from here...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is what i got after integrating \[[4x 2x ^{2}+ 1/3 x ^{3}2/3x ^{3/2}] [2/3x ^{3/2}4x+2x ^{2}1/3x ^{3}]\]

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4that is for the first interval, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, it's for both. ahhhhhhh it's supposed to be + not 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sorry if it takes so long. sorry. you dont need to solve it though. I'll try to do it myself. :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4I am showing both at once here I made a typo above actually, fixed it here\[\int_{0}^{1}f(x)g(x)dx+\int_{1}^{3}g(x)f(x)dx\]\[=\int_{0}^{1}(2x)^2x^{1/2}dx+\int_{1}^{3}x^{1/2}(2x)^2dx\]\[=\int_{0}^{1}44x+x^2x^{1/2}dx+\int_{1}^{3}x^{1/2}4+4xx^2dx\]\[=4x2x^2+\frac{1}{3}x^3\frac{2}{3}x^{3/2}_{0}^{1}+\frac{2}{3}x^{3/2}4x+2x^2\frac{1}{3}x^3_{1}^{3}\]\[=(42+\frac 1 3\frac 2 3)+[\frac2 3(3\sqrt3)4(3)+2(9)][\frac2 3 4+2\frac1 3]\]\[=(2\frac1 3)+2\sqrt3+6(\frac1 32)=10\frac2 3+2\sqrt3\]I don't think that is the answer you wanted, so I must have made an arithmetic mistake somewhere. This is why I usually don't do the definite integration myself. Unless I happen to spot my mistake it will be your job to find it. I know my formulation is correct.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4Oh I found the typo I think...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4I dropped a term at the end\[\int_{0}^{1}f(x)g(x)dx+\int_{1}^{3}g(x)f(x)dx\]\[=\int_{0}^{1}(2x)^2x^{1/2}dx+\int_{1}^{3}x^{1/2}(2x)^2dx\]\[=\int_{0}^{1}44x+x^2x^{1/2}dx+\int_{1}^{3}x^{1/2}4+4xx^2dx\]\[=4x2x^2+\frac{1}{3}x^3\frac{2}{3}x^{3/2}_{0}^{1}+\frac{2}{3}x^{3/2}4x+2x^2\frac{1}{3}x^3_{1}^{3}\]\[=(42+\frac 1 3\frac 2 3)+[\frac2 3(3\sqrt3)4(3)+2(9)9][\frac2 3 4+2\frac1 3]\]\[=(2\frac1 3)+2\sqrt33(\frac1 32)=1\frac2 3+2\sqrt3\]which is the answer stated by your teacher. You got lucky I found it, I was gonna make you do it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoa you're a genius. thanks a lot. i'm trying to find my mistake here. thank you thank you

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.4please do look for it, it's all about practice, I'm no genius your welcome, good luck :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeyyyyyyyyyyyyyyyyyyyyyy i got it. Thanks TurningTest
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