What is the average value of the function # f(x)=e^(-x)*sin(x)# on the interval #[1, pi]#?

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Answer 1 · 2016-03-22T13:04:16

I get #1/(pi-1) int_1^pi e^-x * sinx dx = (e+e^pi(sin(1)+cos(1)))/(2(pi-1)e^(pi+1))#

From the definition of average value of a function, we need to find

#1/(pi-1) int_1^pi e^-x * sinx dx#.

Evaluate #int e^-x * sinx dx# by parts.

Use #u = sinx# and #dv = e^-x dx# (Or the other way around. In this case, either will work.)

We get #du = cosx dx# and #v = -e^-x#.

#uv-int v du = -e^-xsinx + int e^-xcosxdx#

Repeat parts with #u = cosx# and #dv = e^-x dx# for this integral to get

#int e^-x * sinx dx = -e^-xsinx - e^-x-inte^-xsinx dx#.

Add the integral to both sides and divide by #2# then factor, to finish with

#int e^-x sinx dx = -1/(2e^x)(sinx+cosx)#.

Evaluating from #1# to #pi#, we get:

#int_1^pi e^-x sinx dx = -1/(2e^x)(sinx+cosx)]_1^pi#

# = 1/(2e^pi) + (sin(1)+cos(1))/(2e)#

# = (e+e^pi (sin(1)+cos(1)))/(2e^(pi+1))#.

Finally, divide by the length of the interval, to answer:

#1/(pi-1) int_1^pi e^-x sinx dx = (e+e^pi(sin(1)+cos(1)))/(2(pi-1)e^(pi+1))#.