Let f(x) be an even function & I1=∫∞0f(x)dx, I2=∫∞0f(3x−12x)dx, then the value of I1I2 is (where I1&I2 are finite)
Open in App
Solution
3x−12x=tx=16(t+√t2+6),x>0dx=16(1+t√t2+6)dtI2=∫∞−∞f(3x−12x)dx=16∫∞−∞f(t)(1+t√t2+6)dt=16∫∞−∞f(t)dt+16∫∞−∞tf(t)√t2+6dt where first integrand is an even function and second is an odd function. ∴I2=26∫∞0f(t)dt⇒I2=13I1⇒I1I2=3