What is #int_(-ipi)^(ipi) xe^xdx #?

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Answer 1 · 2018-04-10T10:25:30

#int_(-ipi)^(ipi) xe^xdx = -ipi+1 -ipi-1 = -2ipi#

Solve the indefinite integral by parts:

#int xe^xdx = int xd(e^x)#

#int xe^xdx = xe^x - int e^xdx#

#int xe^xdx = xe^x - e^x +C#

#int xe^xdx = e^x(x -1) +C#

Then:

#int_(-ipi)^(ipi) xe^xdx = [e^x(x -1) ]_(-ipi)^(ipi)#

#int_(-ipi)^(ipi) xe^xdx = e^(ipi)(ipi-1) - e^(-ipi)(-ipi-1)#

Using Euler's formula:

#e^(ipi) = cospi + i sinpi = -1#

#e^(-ipi) = cospi - i sinpi = -1#

so:

#int_(-ipi)^(ipi) xe^xdx = -ipi+1 -ipi-1 = -2ipi#