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Math Help - Integral word problem

  1. #1
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    Integral word problem

    I have attempted to answer this question but come up with a number that just doesn't seem correct....please review my solution and let me know where I went wrong. Thanks!

    Find f(2) if f(1/3) = 2 and f ' (x) = e^{3x+2}-3x

    my solution:
    = \frac{1}{3}e^{3x+2}+C

    and since f(1/3) = 2 then: 2 = \frac{1}{3}e^{3(1/3)+2}+C
    and solving for C gives me a not so great number.

    please let me know where I went wrong....thanks!
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  2. #2
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    Quote Originally Posted by lil_cookie View Post
    I have attempted to answer this question but come up with a number that just doesn't seem correct....please review my solution and let me know where I went wrong. Thanks!

    Find f(2) if f(1/3) = 2 and f ' (x) = e^{3x+2}-3x

    my solution:
    = \frac{1}{3}e^{3x+2}+C

    and since f(1/3) = 2 then: 2 = \frac{1}{3}e^{3(1/3)+2}+C
    and solving for C gives me a not so great number.

    please let me know where I went wrong....thanks!
    Hi lil_cookie,

    f'(x)=e^{3x+2}-3x\ \Rightarrow\ f(x)=\frac{1}{3}e^{3x+2}-\frac{3}{2}x^2+C

    Now, C is found from f\left(\frac{1}{3}\right)=2

    Having found C, f(2) can be evaluated.
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  3. #3
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    I understand how you got your answer but the answers I get solving for C and 2 still don't seem right....but I guess working with 'e' is usually like that.

    Thank you
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  4. #4
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    You just need to practice lil_cookie,

    f\left(\frac{1}{3}\right)=e^{\frac{3}{3}+2}-\frac{3}{2}\left(\frac{1}{3}\right)^2+C=e^3-\frac{1}{6}+C=2

    C=\frac{13}{6}-e^3

    Then

    f(2)=e^{6+2}-\frac{3}{2}2^2+\frac{13}{6}-e^3=e^8-6+\frac{13}{6}-e^3
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  5. #5
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    oh you don't actually solve for e



    thank you
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  6. #6
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    Hi lil_cookie,

    you can use the calculator button to evaluate powers of "e".

    e=2.71828 or so.

    The final numerical value of f(2) is e^8-e^3-\frac{36}{6}+\frac{13}{6}

    =2980.957987-20.085537-\frac{23}{6}=2957.04.
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