$$I = \int_{0}^{1} \frac{1}{x^{3/4}} \, \, dx$$ I am facing to get the exact answer of this integral.
My Approach :- Since it is a Improper Definite Integral and function $\frac{1} {x^{\frac{3}{4}}}$ is not defined at x = 0 . So , I can write it as :-
$$
I = \lim_{N\rightarrow 0^{+}}\int_{N}^{1} x^{-3/4} \,dx
= \lim_{N\rightarrow 0^{+} } 4[1^{1/4}- N^{1/4}]
$$
Now , what value should I have to put for $1^{1/4}$?
Since, $1^{1/4} \in \{1,-1 , i , -i\}$. By putting these values , I will get different answers for the same integral but it should have to give a finite single value for this convergent integral which denotes the amount of finite area under the curve within the given limits. I think , I should have to take real values for the real-valued integrand but here I have 2 real values ie. +1 and -1. So, which real value should I have to put for $1^{1/4}$ ? Please help.
Your approach to this problem was based upon finding a primitive of the fnction $x\mapsto x^{-\frac34}$. The primitive that you used is $x\mapsto4\sqrt[4]x$, which maps $1$ into $1$.