What is the antiderivative of #x^2#?

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Answer 1 · 2018-05-23T17:05:30

It is #x^3/3+C#

Note that #intx^ndx=x^(n+1)/(n+1)+C# if #n\ne -1#

Answer 2 · 2018-05-24T04:18:15

#1/3x^3+C#

The antiderivative of a function is the integral of that function. So here we have:

#intx^2 \ dx#

Using the rule that #intx^n \ dx=(x^(n+1))/(n+1)+C,n!=-1#, we find that it equals:

#=(x^(2+1))/(2+1)+C#

#=1/3x^3+C#