
What is the antiderivative of ${{x}^{2}}?$
Answer
531.3k+ views
Hint: We know that the derivative and the integral are both the inverse operations of each other. So, the antiderivative of a function is the integral of that function. We will find the integral of the given function.
Complete step-by-step solution:
Let us consider the given function ${{x}^{2}}$
We are asked to find the antiderivative of the above function. We know that the antiderivative of a function is the integral of that function.
So, we need to find the integral of the function for we are asked to find the antiderivative.
Now, let us recall a familiar basic identity we have learnt so that we can use it to find the integral or antiderivative of the given function.
The identity is given by $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ where $C$ is the constant of integration.
Now, in our case, when we consider the identity, we will get $n=2.$
So, naturally, we will get $n+1=3.$
And from this, we can easily find the integral of the given function.
We will get $\int{{{x}^{2}}dx}=\dfrac{{{x}^{2+1}}}{2+1}+C.$
And therefore, we will get $\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C.$
Hence the antiderivative of the given function ${{x}^{2}}$ is \[\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C.\]
Note: We are asked to find the antiderivative of the given function. So, we can use an alternative method to find the antiderivative by using the meaning that the terminology antiderivative provides. We need to find the antiderivative ${{x}^{2}}.$ So, we need to find the function whose derivative is ${{x}^{2}}.$ We know that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}.$ So, we will get $\dfrac{1}{n}\dfrac{d{{x}^{n}}}{dx}={{x}^{n-1}}.$ In this case, we will get ${{x}^{n-1}}={{x}^{2}},$ which implies $n-1=2$ and therefore, $n=3.$ So we will get $\dfrac{1}{n}\dfrac{d{{x}^{n}}}{dx}=\dfrac{1}{3}\dfrac{d{{x}^{3}}}{dx}=\dfrac{d}{dx}\left( \dfrac{{{x}^{3}}}{3} \right).$ So, if derivative of $\dfrac{{{x}^{3}}}{3}$ is ${{x}^{2}},$ then the antiderivative of \[{{x}^{2}}\] is $\dfrac{{{x}^{3}}}{3}$ and for the indefinite integral we add the constant of integration.
Complete step-by-step solution:
Let us consider the given function ${{x}^{2}}$
We are asked to find the antiderivative of the above function. We know that the antiderivative of a function is the integral of that function.
So, we need to find the integral of the function for we are asked to find the antiderivative.
Now, let us recall a familiar basic identity we have learnt so that we can use it to find the integral or antiderivative of the given function.
The identity is given by $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ where $C$ is the constant of integration.
Now, in our case, when we consider the identity, we will get $n=2.$
So, naturally, we will get $n+1=3.$
And from this, we can easily find the integral of the given function.
We will get $\int{{{x}^{2}}dx}=\dfrac{{{x}^{2+1}}}{2+1}+C.$
And therefore, we will get $\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C.$
Hence the antiderivative of the given function ${{x}^{2}}$ is \[\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C.\]
Note: We are asked to find the antiderivative of the given function. So, we can use an alternative method to find the antiderivative by using the meaning that the terminology antiderivative provides. We need to find the antiderivative ${{x}^{2}}.$ So, we need to find the function whose derivative is ${{x}^{2}}.$ We know that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}.$ So, we will get $\dfrac{1}{n}\dfrac{d{{x}^{n}}}{dx}={{x}^{n-1}}.$ In this case, we will get ${{x}^{n-1}}={{x}^{2}},$ which implies $n-1=2$ and therefore, $n=3.$ So we will get $\dfrac{1}{n}\dfrac{d{{x}^{n}}}{dx}=\dfrac{1}{3}\dfrac{d{{x}^{3}}}{dx}=\dfrac{d}{dx}\left( \dfrac{{{x}^{3}}}{3} \right).$ So, if derivative of $\dfrac{{{x}^{3}}}{3}$ is ${{x}^{2}},$ then the antiderivative of \[{{x}^{2}}\] is $\dfrac{{{x}^{3}}}{3}$ and for the indefinite integral we add the constant of integration.
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