Calculus Integration Help?!

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1 Answer
Mar 27, 2018

See below.

Explanation:

int_0^a omegax cos(omega x) dx =1/omega int_0^a (omega x) cos(omega x)d(omega x)a0ωxcos(ωx)dx=1ωa0(ωx)cos(ωx)d(ωx)

now making y = omega xy=ωx

int_0^a omegax cos(omega x) dx =1/omega int_0^(a omega) y cosy dya0ωxcos(ωx)dx=1ωaω0ycosydy

and the Riemann integral is

lim_(n->0)1/omega sum_(k=1)^n ((a omega)k/n)cos((a omega)(k/n))(a omega)/n

then finally

int_0^(pi/2) 3x cos(3x) dx equiv lim_(n->0)1/3 sum_(k=1)^n ((3pi/2)k/n)cos((3pi/2)(k/n))(3pi/2)/n = -(1/3+pi/2)