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Q.

 Let f(x) be an even function such that 0f(x)dx=π2, then the value of  the definite integral 0fx1xdx is equal to 

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a

π

b

2π

c

3π2

d

π2

answer is B.

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Detailed Solution

 Let I=0fx1xdx..(1) Put x=1tdx=1t2dtFrom(1):I=0f1ttdtt2  Use abf(x)dx=baf(x)dxI=0f1ttdtt2I=0ft1tdtt2 (f is even function )I=0fx1xdxx2...(2)

Adding equations (1) & (2) 

2I=0fx1xdx+0fx1xdxx22I=0fx1x1+1x2dx Put x1x=u1+1x2dx=du

2I=f(u)du  If f is even aaf(x)=2×0af(x)dx=20f(x)dx=2π22I=πI=π2

 

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