Loading [MathJax]/jax/output/HTML-CSS/jax.js


1)

The value of the integral 

π/2π/2(x2+logπ4π+4)cosxdx is 


A) 0

B) π224

C) π22+4

D) π22

Answer:

Option B

Explanation:

π/2π/2(x2+logπxπ+x)cosxdx

 As   aaf(x)dx=0

 when    f(-x) =-f(x)

   I=π/2π/2x2cosxdx+0

 = =2π/20(x2cosx)dx

=2{(x2sinx)π/20π/202x.sinxdx}

=2[π242{xcosx)π/20π/201.(cosx)dx}]

 =2[π242(sinx)π/20]

=2[π242]=(π224)