What is the antiderivative of sqrtx?

2 Answers
Jun 6, 2015

One law of exponentials states that a^(m/n)=root(n)(a^m)

Thus, we can rewrite sqrt(x) as x^(1/2)

Derivating it using the product rule, which states y=a^n, thus y'=n*a^(n-1), we get:

(dy)/(dx)=x^(1/2-1)=x^-(1/2)

However, as another law of exponentials states, a^-n=1/a^n. Thus,

(dy)/(dx)=1/x^(1/2)=1/sqrt(x)

Jun 6, 2015

You can write the square root as:
sqrt(x)=x^(1/2)
Now integrate this using the fact that:
intx^ndx=(x^(n+1))/(n+1)+c
Getting:
intx^(1/2)dx=(x^(1/2+1))/(1/2+1)+c=
=(x^(3/2))/(3/2)+c=2/3(xsqrt(x))+c