calculus

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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 -

    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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