Question

How would I find the average value of f(x)=xsin(x^2) on the interval [sqrt(π)/2,sqrt(π)]?

Answer 1

Umm so if I remember correctly, for average value we'll use:\[\Large\rm f_{ave}(x)=\frac{1}{b-a}\int\limits_a^b f(x)~dx\]So you're taking the area under the curve.
Then think of it as a rectangle (with average height)
Then you divide by the width of the rectangle, giving you this average height (function value).

Answer 2

\[\Large\rm f_{ave}(x)=\frac{1}{\sqrt \pi-\frac{\sqrt \pi}{2}}\int\limits_{\sqrt \pi/2}^{\sqrt \pi} x ~sinx^2~ dx\]

Answer 3

So you need help evaluating this? :o

Answer 4

Yes please.

Answer 5

So let's clean up the fraction in front:
\[\Large\rm f_{ave}(x)=\frac{2}{\sqrt \pi}\int\limits_{\sqrt \pi/2}^{\sqrt \pi} ~sinx^2(x~ dx)\]
Then continue with a u-sub,\[\Large\rm u=x^2\]

Answer 6

It should clean up your integral and also your boundaries really nicely.

Answer 7

So our new boundaries are,\[\Large\rm u\in\left[\frac{\pi}{4},~ \pi\right]\]Understand how I got those?

Answer 8

Yes, I understand.

Answer 9

Understand how to apply the u-sub? +_+
Do ittttt! :D\[\Large\rm f_{ave}(u)=\frac{2}{\sqrt \pi}\int\limits\limits_{u=\pi/4}^{\pi} ~\sin u(x~ dx)\]Gotta change out the x dx portion also.