Asked by m
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?
interval [0; 2]. Compute mA.
what is lim mA as A goes to infinite?
Answers
Answered by
MathMate
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->∞.
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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