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Math Help - What is the integral of this function? I think my answer is wrong

  1. #1
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    What is the integral of this function? I think my answer is wrong

    What is the integral of this?
    dy/dx = /int(5.6 - (e ^(1.2x) + e ^(-1.2x))/2

    I tried it and got the following, I think I am wrong...
    y = 5.6x - 0.6(e^ (1.2x) - e ^(-1.2x)) + c

    Any help would be appreciated

    Thanks!
    Last edited by auonline; July 29th 2009 at 01:01 AM.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by auonline View Post
    What is the integral of this?
    dy/dx = /int(5.6 - (e ^(1.2x) + e ^(-1.2x))/2

    I tried it and got the following, I think I am wrong...
    y = 5.6x - 0.6(e^ (1.2x) - e ^(-1.2x)) + c

    Any help would be appreciated

    Thanks!
    Do you mean this?

    \frac{{dy}}{{dx}} = \frac{1}{2}\left( {5.6 - \left( {{e^{1.2x}} + {e^{ - 1.2x}}} \right)} \right) \Leftrightarrow

    \Leftrightarrow dy = \frac{1}<br />
{2}\left( {5.6 - \left( {{e^{1.2x}} + {e^{ - 1.2x}}} \right)} \right)dx \Leftrightarrow

    \Leftrightarrow \int {dy}  = \frac{1}<br />
{2}\int {\left( {5.6 - {e^{1.2x}} - {e^{ - 1.2x}}} \right)dx}  \Leftrightarrow

     \Leftrightarrow \int {dy}  = \frac{1}{2}\int {\left( {\frac{{28}}{5} - {e^{\left( {{6 \mathord{\left/{\vphantom {6 5}} \right.\kern-\nulldelimiterspace} 5}} \right)x}} - {e^{ - \left( {{6 \mathord{\left/<br />
 {\vphantom {6 5}} \right.\kern-\nulldelimiterspace} 5}} \right)x}}} \right)dx}  \Leftrightarrow

    \Leftrightarrow y = \frac{1}{2}\left( {\frac{{28}}{5}x - \frac{5}<br />
{6}{e^{\left( {{6 \mathord{\left/{\vphantom {6 5}} \right.<br />
 \kern-\nulldelimiterspace} 5}} \right)x}} + \frac{5}{6}{e^{ - \left( {{6 \mathord{\left/{\vphantom {6 5}} \right.\kern-\nulldelimiterspace} 5}} \right)x}}} \right) + C.
    Last edited by DeMath; July 29th 2009 at 01:32 AM. Reason: typo
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  3. #3
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    A little correction : 5.6 = 28/5 ^^
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  4. #4
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    sorry I can't use latex that well it is 5.6 - (e^1.2x + e^-1.2x)/2 \



    EDIT: Nevermind I got it
    Last edited by auonline; July 29th 2009 at 01:56 AM.
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