The second one is correct, because the arbitrary constant is associated with the integral. Since the x on the outside multiplies the entire integral, it must also multiply the constant.
The second one is correct, because the arbitrary constant is associated with the integral. Since the x on the outside multiplies the entire integral, it must also multiply the constant.
Thanks. But if the second one is correct then we know that if two integrals are equal they are differed by a constant. so LHS and RHS are differed by a constant C. So shouldn't
be right also? Am i interpreting the theory right?
Last edited by x3bnm; November 8th 2010 at 09:16 AM.
Because the variable is outside the integral that is why x is multiplied with the whole integral including the constant. Sorry for posting before reading the reply carefully. Thanks Ackbeet.