Showing that these two integrals are equal

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SUMMARY

The discussion centers on demonstrating the equality of two integrals involving the functions \( e^x \) and \( e^{-x} \). Participants emphasize the importance of evaluating both integrals to understand the factor of 2 that arises from the symmetry of the function \( f(x) = e^x + e^{-x} \). The equality holds due to the property that if \( f(x) = f(-x) \), then \( \int_{-a}^a f(x) \, dx = 2\int_0^a f(x) \, dx \). This principle is fundamental in calculus and is applicable to similar integrals.

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conv
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These two are equal to each other, but I can't figure out how they can be that.

1604049271124.png


I know that 2 can be taken out if its in the function, but where does the 2 come from here?
 
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Why not evaluate the integrals to check they are the same?

If you do that you might see for yourself where the ##2## comes from.
 
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so do i solve for the integral on the left side and end up with the integral on the right?? Or do i solve them just as they are written above? (I have solved the equation with values, but don't really know how to do this one. )
 
conv said:
so do i solve for the integral on the left side and end up with the integral on the right?? Or do i solve them just as they are written above?(I have solved the equation with values, but don't really know how to do this one. )
I suggest you evaluate both integrals. And check they are equal.

I assume you know how to integrate ##e^x## and ##e^{-x}##?
 
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i understood it. Thanks
 
conv said:
i understood it. Thanks

More generally, if ##f(x) = f(-x)## then ##\int_{-a}^a f(x) = 2\int_0^a f(x)##. On the other hand, if ##f(x) = -f(-x)## then ##\int_{-a}^a f(x) = 0##. Can you see why ##f(x) = e^x + e^{-x}## satisfies ##f(x) = f(-x)##?
 
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This should be almost self-evident and not difficult to argue from the symmetry around 0 between the two functions ##e^x## and ##e^{-x}##. If in doubt about that, plot them. Which would only be reminding yourself of what you should securely know already. Not clever but rather basic.

ETA OK this is equivalent to what etotheipi says.
 

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