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Math Help - confused

  1. #1
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    confused

    If  f'(x)=cos(x^2-1) and f(-1)=1.5, then f(5)= ?

    so far i have:

     \int_{-1}^5 cos(x^2-a) dx= f(5)-f(-1)

    i let  u=x^2-1
     du=2x

    change the limits from x=5 to u=24, x=-1 to u=0

    and have
     1/2 \int_0^{24} cos(u) du + f(-1)

    which gives me -1.788 but that isn't any of the answers provided!
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  2. #2
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    Quote Originally Posted by frog09 View Post
    If  f'(x)= cos(x^2-1) and f(-1)=1.5, then f(5)= ?

    so far i have:

     \int_{-1}^5 cos(x^2-a) dx= f(5)-f(-1)

    i let  u=x^2-1
     du=2x Mr F says: du = 2x dx =>dx = du/(2x). This will lead nowhere for you because the x will cause trouble. The integral cannot in fact be found using a finite number of elementary functions.

    change the limits from x=5 to u=24, x=-1 to u=0 and have

     1/2 \int_0^{24} cos(u) du + f(-1) Mr F says: Wrong. What happened to the 1/x part of du ....?

    which gives me -1.788 but that isn't any of the answers provided!
     \int_{-1}^5 \cos(x^2-1) \, dx = f(5) - f(-1)

    \Rightarrow f(5) = \int_{-1}^5 \cos(x^2-1) \, dx + f(-1)

     = \int_{-1}^5 \cos(x^2-1) \, dx + 1.5

    = 1.5244 + 1.5 correct to four decimal places using my TI-89

    = 3.0244.
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  3. #3
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    i don't know what i am doing wrong with this portion of the problem:
     \int_{-1}^5 cos(x^2-1) dx + 1.5

    i keep getting by antiderving cos
     \int_{-1}^5 sin(x^2-1)

    and when i evaluate that between -1 and 5
     sin(5^2-1)-sin({-1}^2-1)=sin(24)-sin(0)
    i get -.9056 which when added to 1.5 isn't correct. could you guide me as to where i went awry? thanks. =/
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  4. #4
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    Quote Originally Posted by frog09 View Post
    i don't know what i am doing wrong with this portion of the problem:
     \int_{-1}^5 cos(x^2-1) dx + 1.5

    i keep getting by antiderving cos
     \int_{-1}^5 sin(x^2-1)

    and when i evaluate that between -1 and 5
     sin(5^2-1)-sin({-1}^2-1)=sin(24)-sin(0)
    i get -.9056 which when added to 1.5 isn't correct. could you guide me as to where i went awry? thanks. =/
    I have already told you in my earlier reply that \int \cos(x^2-1) \, dx cannot be found in terms of a finite number of elementary functions. I don't know why you're insisting on trying to integrate it.

    \int \cos(x^2-1) \, dx \neq \sin (x^2 - 1) .... if you bother to differentiate \sin (x^2 - 1) (using the chain rule) you will quickly realise this.
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