calculus
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 20=2.
I=∫sin(Ax)^3dx
Substitute
u=cos(Ax),
du =Asin(Ax)dx
so
I=(1/A)∫(1u²)du
=(1/A)[uu³/3] from 0 to 2
=(cos^3(2A)cos^3(0)3cos(2A)+3cos(0))/3A
=(cos^3(2A)3cos(2A)+2)/3A
mA = I/2 = (cos^3(2A)3cos(2A)+2)/6A
For the second part, take the limit as A>∞.