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I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:

$ \int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x = \int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x $

$ = \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6 = \left[ \frac{8e^{4x} + 16e^{2x} + 8x}{e^{2x}} \right]_{-6}^6 $

$ = (\frac{8e^{24} + 16e^{12} + 48}{e^{12}}) - (\frac{8e^{-24} + 16e^{-12} - 48}{e^{-12}}) $

$ = e^{-12}(8e^{24} + 16e^{12} + 48) - e^{12}(8e^{-24} + 16e^{-12} - 48) $

$ = 8e^{12} + 16 + 48e^{-12} - (8e^{-12} + 16 - 48e^{12}) $

$ = 8e^{12} + 16 + 48e^{-12} - 8e^{-12} - 16 + 48e^{12}) $

$ = 56e^{12} + 56e^{-12} $

Where am I going wrong?

  • 0
    Use $\frac{1}{e^{2x}}=e^{-2x}$ in the simplication step before integrating.2012-07-21

4 Answers 4

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$ \int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x= \int_{-6}^6 \frac{16e^{4x} + 16e^{2x}+ 4}{e^{2x}} \, \mathrm{d} x= \int_{-6}^6 16e^{2x} + 16+ 4e^{-2x} \, \mathrm{d} x= \left[ 8e^{2x} + 16x-2e^{-2x} \right]_{-6}^6= 8(e^{12}-e^{-12}) + 16\cdot 12 -2(e^{-12}-e^{12})= 192+ 10 e^{12}-10 e^{-12} $

You can check both indefinite and definite integral at WolframAlpha.


I am not sure where is mistake in your solution (since I do not understand what exactly you have done), but most probably you have used $\int \frac{f(x)}{g(x)} \, \mathrm{d} x = \frac{\int f(x) \, \mathrm{d} x}{\int g(x)\, \mathrm{d} x}$, as suggested by Gerry's comment. This formula is incorrect.

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$ I:=\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$

Let $u=e^x, du = e^x \ dx$, leaving us with:

$\int_{e^{-6}}^{e^{6}} \frac{\left( 4u^2 + 2 \right)^2}{u^3} \ du$

Expand the numerator to get

$\int_{e^{-6}}^{e^{6}} \frac{16u^4 + 16u^2 + 4}{u^3} \ du$

Since the highest power in the numerator is greater than the highest power in the denominator, we have to do some long division. Upon dividing, you get:

$\int_{e^{-6}}^{e^{6}} \frac{4}{u^3} + 16u + \frac{16}{u} \ du$

Integrate to get:

$8u^2-\frac{2}{u^2} + 16 \ln |u|$

Back-substitute $u=e^x$ to get

$8e^{2x} - 2e^{-2x} + 16 \ln|e^{x}|$

Since $e^x$ is strictly increasing, we can drop the absolute value. Also, recall that $\ln{e^x} = x$, so you can simplify a bit.

$8e^{2x} + 16x - 2e^{-2x}$

Now, simply evaluate at your endpoints to find that

$I \approx 1.628\times10^6$

  • 0
    @JavierBadia: Yes. I missed that one, excellent observation!2012-07-22
3

You had these steps ok: $ \int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x = \int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x $

After that, there are a number of choices. It looks like you forgot to integrate the solution.

You could do this: $\int_{-6}^6 {\frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} dx}$ $= \int_{-6}^6 { \left( 16e^{2x} + 16 + 4e^{-2x} \right) dx}$ $= \left[ { 8e^{2x} + 16x - 2e^{-2x} } \right]_{-6}^6$

The integration is directly above. Plugging in the values then gives:

$ \left( 8e^{2(6)} + 16(6) - 2e^{-2(6)} \right) - \left( 8e^{2(-6)} + 16(-6) - 2e^{-2(-6)} \right) $ $= \left( 8e^{12} + 96 - 2e^{-12} \right) - \left( 8e^{-12} -96 - 2e^{12} \right) $ $= 10e^{12} + 192 - 10e^{-12} $

$\approx 1.62774*10^6$

To get the hyperbolic sine ($\sinh$), note that $ \sinh(x) = \frac{ e^{x} - e^{-2x} } {2}$ $ \sinh(12) = \frac{ e^{12} - e^{-12} } {2}$ $20\sinh(12) = 10 \left( e^{12} - e^{-12} \right)$

So we have $ 10e^{12} - 10e^{-12} + 192 $ $= 20\sinh(12) + 192 $ $= 4 \left( 5 \sinh(12) + 48 \right)$

  • 0
    Note that this answer is also equal to $4 (48 + 5 \sinh(12))$2012-07-21
2

As for the error in your work, I see a problem in the following step:

$ \int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x = \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6$

The denominator is not a constant, so you cannot do the integration like this. I would suggest dividing the numerator by the denominator. This amounts to the substitution which Joe suggests, but seems less complicated in my opinion.

Also, the 2 outside the parenteses in the numerator is incorrect.

  • 0
    @GerryMyerson Ahhh...that makes sense, even if it is incorrect.2012-07-22