Calculus Integration Help?!

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1 Answer
Mar 27, 2018

Answer:

See below.

Explanation:

#int_0^a omegax cos(omega x) dx =1/omega int_0^a (omega x) cos(omega x)d(omega x)#

now making #y = omega x#

#int_0^a omegax cos(omega x) dx =1/omega int_0^(a omega) y cosy dy#

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)#