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Math Help - integral

  1. #1
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    integral

    consider the function f(x) whose second derivative is f''(x) = 3x+2sin(x)
    if f(0)=4, f'(x)=4. what is f(2) = ?

    heres what i did
     f'(x) = \int (3x+2sin(x)
     = \frac{3x^2}{2} - 2cos(x) + C1

     f(x) = \int \frac{3x^2}{2} - 2cos(x) + C1
     = \frac{x^3}{2} + C2 - 2sin(x) + C2
    Last edited by viet; November 11th 2006 at 09:19 PM. Reason: typo
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  2. #2
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    i thik something is wrong how can a function which has sin o cos i it give the derivative as 4 that is a constant?
    it ought to be a straight line for that
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  3. #3
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    Quote Originally Posted by srinivas View Post
    i thik something is wrong how can a function which has sin o cos i it give the derivative as 4 that is a constant?
    it ought to be a straight line for that
    Because it is a typo in the intial conditions. Presumably they shout be:

    f(0)=4, f'(0)=4.

    RonL
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  4. #4
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    Quote Originally Posted by viet View Post
    consider the function f(x) whose second derivative is f''(x) = 3x+2sin(x)
    if f(0)=4, f'(x)=4. what is f(2) = ?

    heres what i did
     f'(x) = \int (3x+2sin(x)
     = \frac{3x^2}{2} - 2cos(x) + C1

     f(x) = \int \frac{3x^2}{2} - 2cos(x) + C1
     = \frac{x^3}{2} + C2 - 2sin(x) + C2
    Last line should read:

     = \frac{x^3}{2} - 2\sin(x) + C1\ x + C2

    RonL
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  5. #5
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    Hello, viet!

    I agree with the Captain: there must be a typo in the problem.
    I'll assume it is supposed to be: f'(0) = 4


    Consider the function f(x) whose second derivative is: f''(x) \:= \:3x+2\sin x
    If f(0) = 4,\;f'(0) = 4, find f(2)

    We have: . f'(x)\;=\;\int \left(3x + 2\sin x\right)\,dx \;= \;\frac{3}{2}x^2 - 2\cos x + C_1

    Since f'(0) = 4, we have: . \frac{3}{2}(0^2) - 2\cos0 + C_1\:=\:4\quad\Rightarrow\quad C_1 = 6

    . . Hence: . f'(x)\:=\:\frac{3}{2}x^2 - 2\cos x + 6


    We have: . f(x)\;=\;\int\left(\frac{3}{2}x^2 - 2\cos x + 6\right)\,dx \;=\;\frac{1}{2}x^3 - 2\sin x + 6x + C_2

    Since f(0) = 4, we have: . \frac{1}{2}(0^3) - 2\sin0 + 6(0) + C_2\:=\:4\quad\Rightarrow\quad C_2 = 4

    . . Hence: . \boxed{f(x) \;= \;\frac{1}{2}x^3 - 2\sin x + 6x + 4}


    Therefore: . f(2) \;= \;\frac{1}{2}(2^3) = 2\sin2 + 6(2) + 4 . . . etc.

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