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Question

Let f(x) be an even function & I1=0f(x) dx,
I2=0f(3x12x) dx , then the value of I1I2 is (where I1 & I2 are finite)

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Solution

3x12x=tx=16(t+t2+6),x>0dx=16(1+tt2+6)dtI2=f(3x12x)dx=16f(t)(1+tt2+6) dt=16f(t) dt+16tf(t)t2+6 dt
where first integrand is an even function and second is an odd function.
I2=260f(t) dtI2=13I1I1I2=3

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