Question

What is the antiderivative of `sqrtx`?

Answer 1

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)`

Answer 2

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`