Posted by **m** on Wednesday, February 23, 2011 at 3:00pm.

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?

- calculus -
**MathMate**, Wednesday, February 23, 2011 at 7:21pm
mA is the integral divided by the length of the interval, namely 2-0=2.

I=∫sin(Ax)^3dx

Substitute

u=cos(Ax),

du =-Asin(Ax)dx

so

I=-(1/A)∫(1-u²)du

=-(1/A)[u-u³/3] 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->∞.

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