Question

integrate mod(xcos pix) limit 0 to 3/2

Answer 1

Answer:

0

Step-by-step explanation:

Given integrate mod(xcos pix) limit 0 to 3/2

Given to integrate 0 to 3/2 mod x cosπx dx

    = x∫_0^(3/2)▒cosπxdx -∫_0^(3/2)▒〖d/dx(x)〗 (∫_0^(3/2)▒cosπxdx

= [x (sinπx/π) 0 to 3/2 - ∫_0^(3/2)▒1[(sin3π/2)/π-0]dx

= [3/2(sin3π/2/π) – 0] - ∫_0^(3/2)▒1[(sin3π/2)/π-0]dx  

= 3/2π(sin270) – 0 - ∫_0^(3/2)▒〖1/π(sin270) -0dx〗

= 3/2π(-1) – 0 - ∫_0^(3/2)▒〖-1/πdx〗

= -3/2π + 1/π (x) 0 to 3/2

= - 3 / 2π + 3/2π  

∫_0^(3/2)▒xcosπxdx= 0

 

Answer 2

Answer:

Answer of this question is

5π-1/2π∧2