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#