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April 16, 2014

April 16, 2014

Posted by **alsa** on Saturday, June 16, 2012 at 8:56pm.

- Integral calculus -
**MathMate**, Sunday, June 17, 2012 at 6:25pmTo show the results, we assume

F(x)=∫f(x)dx

then

∫f(x)dx from 0 to a is F(a)-F(0)

and

∫f(x-a)dx from 0 to a is F(0)-F(0-a)

=F(0)-F(-a)

=-F(-a)+F(0)

For the two to be equal, we require:

F(a)-F(0) = -F(-a) + F(0)

or

F(a)-F(-a) = 2F(0)

Which is not generally true. So the answer is no. A counter example is when f(x)=sin(x).

However, equality can be satisfied if F(x) is an odd function where F(0)=0 (such as sin(x)). This means that equality will hold if f(x)=±k*cos(x).

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