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

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

Mathematics
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we are going to need the point where these two graphs intersect, because they will cross each other, can you find that?
|dw:1327193397919:dw|we 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\[f(x)=g(x)\]
I 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????

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yes those are the right intervals
Use Riemann sums, Integration is overrated :P
so 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
Great, which one do you like most the right or the left one?
their average :P
hehe :D Can't we use there median? :P
hey 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?
right :)
Turing is a awesome teacher! :)
@turing test.. your graph is not right... f(x)=(2-x)^2 have minima at x=2, and also x=2 is the solution... after that it may not cut squareroot of x before 3
thx, only to good students @sam I really was only trying to illustrate the intersection, there is no scale, so it flies for our purposes
(1,1) is the intersection
do you know how to find that area by calculator?
no, I know how to find it by integrating
how? I can't find the area for [1,3] by calculator
YOu could use Mathematica :)
Turning Test: thanks for your help. :) I'm trying to get the answer. FoolForMath: keke i know right. but it takes so long
\[\int_{1}^{3}g(x)-f(x)dx=\int_{1}^{3}x^{1/2}-(2-x)^2dx=\int_{1}^{3}x^{1/2}-(4-4x+x^2)dx\]got it from here?
\[=\int_{1}^{3}x^{1/2}-(4-4x+x^2)dx=\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]...equals blah blah It's just algebra after integrating
ok I'm confused now ~.~
where did you get stuck? did you understand everything I typed so far?
you didn't type the one from [0,1), did you?
no, 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.
:) 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.
\[\int_{0}^{1}f(x)-g(x)dx+\int_{1}^{3}g(x)-f(x)dx\]\[=\int_{0}^{1}(2-x)^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-(2-x)^2dx\]\[\int_{0}^{1}2-4x+x^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]...
\[=2x-2x^2+\frac{1}{3}x^3-\frac{2}{3}x^{3/2}|_{0}^{1}+\frac{2}{3}x^{3/2}-4x+2x^2-\frac{1}{3}x^3|_{1}^{3}\]\[=(2-2+\frac 1 3-\frac 2 3)+[\frac2 3(3\sqrt3)-4(3)+2(9)]-[\frac2 3 -4+2-\frac1 3]\]...just simplifying from here...
here is what i got after integrating \[[4x- 2x ^{2}+ 1/3 x ^{3}-2/3x ^{3/2}]- [2/3x ^{3/2}-4x+2x ^{2}-1/3x ^{3}]\]
that is for the first interval, right?
no, it's for both. ahhhhhhh it's supposed to be + not -
I'm sorry if it takes so long. sorry. you dont need to solve it though. I'll try to do it myself. :)
I am showing both at once here I made a typo above actually, fixed it here\[\int_{0}^{1}f(x)-g(x)dx+\int_{1}^{3}g(x)-f(x)dx\]\[=\int_{0}^{1}(2-x)^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-(2-x)^2dx\]\[=\int_{0}^{1}4-4x+x^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]\[=4x-2x^2+\frac{1}{3}x^3-\frac{2}{3}x^{3/2}|_{0}^{1}+\frac{2}{3}x^{3/2}-4x+2x^2-\frac{1}{3}x^3|_{1}^{3}\]\[=(4-2+\frac 1 3-\frac 2 3)+[\frac2 3(3\sqrt3)-4(3)+2(9)]-[\frac2 3 -4+2-\frac1 3]\]\[=(2-\frac1 3)+2\sqrt3+6-(\frac1 3-2)=10-\frac2 3+2\sqrt3\]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.
Oh I found the typo I think...
I dropped a term at the end\[\int_{0}^{1}f(x)-g(x)dx+\int_{1}^{3}g(x)-f(x)dx\]\[=\int_{0}^{1}(2-x)^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-(2-x)^2dx\]\[=\int_{0}^{1}4-4x+x^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]\[=4x-2x^2+\frac{1}{3}x^3-\frac{2}{3}x^{3/2}|_{0}^{1}+\frac{2}{3}x^{3/2}-4x+2x^2-\frac{1}{3}x^3|_{1}^{3}\]\[=(4-2+\frac 1 3-\frac 2 3)+[\frac2 3(3\sqrt3)-4(3)+2(9)-9]-[\frac2 3 -4+2-\frac1 3]\]\[=(2-\frac1 3)+2\sqrt3-3-(\frac1 3-2)=1-\frac2 3+2\sqrt3\]which is the answer stated by your teacher. You got lucky I found it, I was gonna make you do it ;)
whoa you're a genius. thanks a lot. i'm trying to find my mistake here. thank you thank you
please do look for it, it's all about practice, I'm no genius your welcome, good luck :D
yeyyyyyyyyyyyyyyyyyyyyyy i got it. Thanks TurningTest
very welcome!

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