`f(t) = e^sin(t) cos(t), [0, pi/2]` Find the average value of the function on the given interval.

The average value is the integral of a function over an interval divided by the length of this interval. The length of interval is `pi/2` here, to find the integral start from the indefinite integral.

Make the substitution `u=sin(t),` then `du=cos(t)dt` and the integral is

`int e^u du=e^u+C=e^(sin(t))+C.`

So the definite integral is

`e^(sin(t))|_(t=0)^(pi/2)=e^1-e^0=e-1.`

And the average is `(2(e-1))/pi.`

The average value is the integral of a function over an interval divided by the length of this interval. The length of interval is `pi/2` here, to find the integral start from the indefinite integral.

Make the substitution `u=sin(t),` then `du=cos(t)dt` and the integral is

`int e^u du=e^u+C=e^(sin(t))+C.`

So the definite integral is

`e^(sin(t))|_(t=0)^(pi/2)=e^1-e^0=e-1.`

And the average is `(2(e-1))/pi.`