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anonymous

  • 5 years ago

integrating from sinx to 1 (t^2 f(t)dt=1-sinx for all xE R ,then f(1/sq.root(3))

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  1. anonymous
    • 5 years ago
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    Try using the Equation button in the bottom.

  2. anonymous
    • 5 years ago
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    Is it \[\int\limits_{\sin x}^{1}t^2f(t) dt=1-\sin x?\]

  3. anonymous
    • 5 years ago
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    Are you there?

  4. anonymous
    • 5 years ago
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    Just tell me if what I wrote there is what you're asking about :)

  5. anonymous
    • 5 years ago
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    yaa

  6. anonymous
    • 5 years ago
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    Good. Do you know the fundamental theorem of calculus?

  7. anonymous
    • 5 years ago
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    i dunn mayb

  8. anonymous
    • 5 years ago
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    Hmm. I will start by taking the derivative of both sides. Taking the derivative of the integral can be done using: \[\int\limits_{a}^{g(x)}f(t)dt=f(g(x)).g'(x)\]

  9. anonymous
    • 5 years ago
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    Sorry write before the integral d/dx

  10. anonymous
    • 5 years ago
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    but y

  11. anonymous
    • 5 years ago
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    oh i gotta

  12. anonymous
    • 5 years ago
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    \[{d \over dx}\int\limits_{\sin x}^{1}t^2f(t)dt={d \over dx}(1-\sin x)\]

  13. anonymous
    • 5 years ago
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    You are leaving?

  14. anonymous
    • 5 years ago
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    no

  15. anonymous
    • 5 years ago
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    You got the formula I wrote above?

  16. anonymous
    • 5 years ago
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    yaa but then with this we hav to find f(\[ 1/\sqrt{3}\])

  17. anonymous
    • 5 years ago
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    I know.. We have to find f(x) first.

  18. anonymous
    • 5 years ago
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    oh

  19. anonymous
    • 5 years ago
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    It will take just two more steps.

  20. anonymous
    • 5 years ago
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    k

  21. anonymous
    • 5 years ago
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    \[\implies -(\sin^2 x)f(\sin x)\cos x=-\cos x \implies \cos x(- \sin^2 x f(\sin x)+1)=0\]

  22. anonymous
    • 5 years ago
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    It follows that: cos x=0 or \[\sin^2 x f(\sin x)=1 \implies f(\sin x)={1 \over \sin^2 x}\]

  23. anonymous
    • 5 years ago
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    Now to find the value of f(1/sqrt(3)).. put x=pi/3

  24. anonymous
    • 5 years ago
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    k

  25. anonymous
    • 5 years ago
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    Sorry.. not pi/3, just substitute x=sin^-1(1/sqrt(3))you will get:

  26. anonymous
    • 5 years ago
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    bot acc to formula g(x) =1 then y we substitute sinx

  27. anonymous
    • 5 years ago
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    \[f(\sin(\sin^{-1}(1/\sqrt3))=f(1/\sqrt3)={1 \over \sin^2(\sin^{-1}{1\over \sqrt3})}={1 \over 1/3}=3\]

  28. anonymous
    • 5 years ago
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    Do you have the answer?

  29. anonymous
    • 5 years ago
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    you know that sin(arcsinx)=x, right? because arcsin just "undoes" the sin since it's its inverse.

  30. anonymous
    • 5 years ago
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    3

  31. anonymous
    • 5 years ago
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    Ok.. that's what I got and showed above. I hope the steps are clear, and make sense to you.

  32. anonymous
    • 5 years ago
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    thnks

  33. anonymous
    • 5 years ago
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    r u teacher

  34. anonymous
    • 5 years ago
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    No. I am a student.

  35. anonymous
    • 5 years ago
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    quit intelligent

  36. anonymous
    • 5 years ago
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    Thanks!! :)

  37. anonymous
    • 5 years ago
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    passed out 12 this yr

  38. anonymous
    • 5 years ago
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    What?

  39. anonymous
    • 5 years ago
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    hav u passed out 12 this yr?

  40. anonymous
    • 5 years ago
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    hav u passed out 12 this yr?

  41. anonymous
    • 5 years ago
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    I don't know what that means.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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