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anonymous

  • 5 years ago

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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  1. TuringTest
    • 5 years ago
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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?

  2. TuringTest
    • 5 years ago
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    |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\[f(x)=g(x)\]

  3. anonymous
    • 5 years ago
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    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????

  4. TuringTest
    • 5 years ago
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    yes those are the right intervals

  5. anonymous
    • 5 years ago
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    Use Riemann sums, Integration is overrated :P

  6. TuringTest
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    Great, which one do you like most the right or the left one?

  8. TuringTest
    • 5 years ago
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    their average :P

  9. anonymous
    • 5 years ago
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    hehe :D Can't we use there median? :P

  10. anonymous
    • 5 years ago
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    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?

  11. TuringTest
    • 5 years ago
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    right :)

  12. anonymous
    • 5 years ago
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    Turing is a awesome teacher! :)

  13. anonymous
    • 5 years ago
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    @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

  14. TuringTest
    • 5 years ago
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    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

  15. TuringTest
    • 5 years ago
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    (1,1) is the intersection

  16. anonymous
    • 5 years ago
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    do you know how to find that area by calculator?

  17. TuringTest
    • 5 years ago
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    no, I know how to find it by integrating

  18. anonymous
    • 5 years ago
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    how? I can't find the area for [1,3] by calculator

  19. anonymous
    • 5 years ago
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    YOu could use Mathematica :)

  20. anonymous
    • 5 years ago
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    Turning Test: thanks for your help. :) I'm trying to get the answer. FoolForMath: keke i know right. but it takes so long

  21. TuringTest
    • 5 years ago
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    \[\int_{1}^{3}g(x)-f(x)dx=\int_{1}^{3}x^{1/2}-(2-x)^2dx=\int_{1}^{3}x^{1/2}-(4-4x+x^2)dx\]got it from here?

  22. TuringTest
    • 5 years ago
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    \[=\int_{1}^{3}x^{1/2}-(4-4x+x^2)dx=\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]...equals blah blah It's just algebra after integrating

  23. anonymous
    • 5 years ago
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    ok I'm confused now ~.~

  24. TuringTest
    • 5 years ago
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    where did you get stuck? did you understand everything I typed so far?

  25. anonymous
    • 5 years ago
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    you didn't type the one from [0,1), did you?

  26. TuringTest
    • 5 years ago
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    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.

  27. anonymous
    • 5 years ago
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    :) 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.

  28. TuringTest
    • 5 years ago
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    \[\int_{0}^{1}f(x)-g(x)dx+\int_{1}^{3}g(x)-f(x)dx\]\[=\int_{0}^{1}(2-x)^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-(2-x)^2dx\]\[\int_{0}^{1}2-4x+x^2-x^{1/2}dx+\int_{1}^{3}x^{1/2}-4+4x-x^2dx\]...

  29. TuringTest
    • 5 years ago
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    \[=2x-2x^2+\frac{1}{3}x^3-\frac{2}{3}x^{3/2}|_{0}^{1}+\frac{2}{3}x^{3/2}-4x+2x^2-\frac{1}{3}x^3|_{1}^{3}\]\[=(2-2+\frac 1 3-\frac 2 3)+[\frac2 3(3\sqrt3)-4(3)+2(9)]-[\frac2 3 -4+2-\frac1 3]\]...just simplifying from here...

  30. anonymous
    • 5 years ago
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    here is what i got after integrating \[[4x- 2x ^{2}+ 1/3 x ^{3}-2/3x ^{3/2}]- [2/3x ^{3/2}-4x+2x ^{2}-1/3x ^{3}]\]

  31. TuringTest
    • 5 years ago
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    that is for the first interval, right?

  32. anonymous
    • 5 years ago
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    no, it's for both. ahhhhhhh it's supposed to be + not -

  33. anonymous
    • 5 years ago
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    I'm sorry if it takes so long. sorry. you dont need to solve it though. I'll try to do it myself. :)

  34. TuringTest
    • 5 years ago
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    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.

  35. TuringTest
    • 5 years ago
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    Oh I found the typo I think...

  36. TuringTest
    • 5 years ago
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    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 ;)

  37. anonymous
    • 5 years ago
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    whoa you're a genius. thanks a lot. i'm trying to find my mistake here. thank you thank you

  38. TuringTest
    • 5 years ago
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    please do look for it, it's all about practice, I'm no genius your welcome, good luck :D

  39. anonymous
    • 5 years ago
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    yeyyyyyyyyyyyyyyyyyyyyyy i got it. Thanks TurningTest

  40. TuringTest
    • 5 years ago
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    very welcome!

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