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Let f be a positive function Let I1=∫k1−kx f{x(1−x)}dx I2=∫k1−kf{x(1−x)}dx where 2k−1>0,. If I2=pI1 then the value of p is

Answer» Let f be a positive function
Let I1=k1kx f{x(1x)}dx
I2=k1kf{x(1x)}dx
where 2k1>0,. If I2=pI1 then the value of p is


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