Posted by m on .
Suppose A is a positive real number and mA is the average value of (sin(Ax))3 on the
interval [0; 2]. Compute mA.
what is lim mA as A goes to infinite?
mA is the integral divided by the length of the interval, namely 2-0=2.
=-(1/A)[u-u³/3] from 0 to 2
mA = I/2 = (cos^3(2A)-3cos(2A)+2)/6A
For the second part, take the limit as A->∞.