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    I have inserted a picture of 2 different ways I have worked the area of a curve out for Y=4e^2x

    Would someone be able to tell me if one of them is right please?
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    (Original post by ak1992)
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    I have inserted a picture of 2 different ways I have worked the area of a curve out for Y=4e^2x

    Would someone be able to tell me if one of them is right please?
    Neither is correct. Use the result that \displaystyle \int e^{ax} \, \mathrm{d}x = \frac{e^{ax}}{a} + c
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    Try to use a bit of care in your notation. You answer shows a very common misconception when dealing with integration of composite functions. In this case we have \int fg(x) where f(x)=e^x\;\text{and}\;g(x)=2x\;\  rightarrow\; fg(x)=e^2x

    The case where g(x) is just of the form  ax is particularly easy to deal with as the differential of  ax is just
     a and needs to divide the result of integrating  f g(x) in a way that treats g(x) as if it were just x .

    Hence \int \cos 2x = \frac{\sin 2x} {2} , \int e^{ 2x} = \frac{e^{2x}} {2} and \int \sec^2(2X) = \frac{\tan 2x} {2}
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    (Original post by ak1992)
    I thought that was only for Intergration?
    And how precisely do you find the area under a curve? By integration.

    The area under a curve between the point x=a and x=b of a function f(x) is given by \displaystyle \int_a^b f(x) \, \mathrm{d}x.

    In your case, the area under the curve of 4e^{2x} between x=1 and x=3 is given by \displaystyle \int_1^3 4e^{2x} \, \mathrm{d}x
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    I think it's important for you to go back and understand the content before continuing with any more questions. Otherwise you're just blindly trying to apply algorithms that you don't understand.
 
 
 
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