
What is the anti – derivative of a constant?
Answer
530.7k+ views
Hint: We solve this problem by using integration because the anti – derivative is another name of integration. We assume some constant and apply the integration to that constant to get the required answer. We use the standard formula of power rule of integration as,
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
Complete step-by-step solution:
We are asked to find the anti – derivative of a constant.
Let us assume the constant function as,
$\Rightarrow f\left( x \right)=C$
Where, $'C'$ is some constant.
We know that anti – derivative is nothing but integration.
Now, let us integrate the above assumed equation with respect to $'x'$ then we get,
$\begin{align}
& \Rightarrow \int{f\left( x \right).dx}=\int{C.dx} \\
& \Rightarrow \int{f\left( x \right).dx}=C\int{1.dx} \\
\end{align}$
We know that the number ‘1’ can be written as anything power ‘0’.
Here, we can see that the integration is done with respect to $'x'$ so, let us represent the number ‘1’ as ${{x}^{0}}$
Now, by substituting the above representation of ‘1’ in the required integral then we get,
$\Rightarrow \int{f\left( x \right).dx}=C\int{{{x}^{0}}.dx}$
We know that the standard formula of power rule of integration is given as,
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
By using this power rule to above integral then we get,
$\begin{align}
& \Rightarrow \int{f\left( x \right).dx}=C\left( \dfrac{{{x}^{0+1}}}{0+1}+c \right) \\
& \Rightarrow \int{f\left( x \right).dx}=Cx+Cc \\
& \Rightarrow \int{f\left( x \right).dx}=Cx+{{C}_{1}} \\
\end{align}$
Where, ${{C}_{1}}$ is some other constant not equal to $C$
Therefore, we can conclude that the required integral also called as anti – derivative of constant as,
$\therefore \int{f\left( x \right).dx}=Cx+{{C}_{1}}$
Note: Students may make mistakes mainly by not taking the constant after the integration.
The power rule we used to solve this problem is given as,
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
Here, we can see that constant $'c'$ is very important in case of finding the equation after the integration. Students may miss this constant some times and gives the answer which is not correct.
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
Complete step-by-step solution:
We are asked to find the anti – derivative of a constant.
Let us assume the constant function as,
$\Rightarrow f\left( x \right)=C$
Where, $'C'$ is some constant.
We know that anti – derivative is nothing but integration.
Now, let us integrate the above assumed equation with respect to $'x'$ then we get,
$\begin{align}
& \Rightarrow \int{f\left( x \right).dx}=\int{C.dx} \\
& \Rightarrow \int{f\left( x \right).dx}=C\int{1.dx} \\
\end{align}$
We know that the number ‘1’ can be written as anything power ‘0’.
Here, we can see that the integration is done with respect to $'x'$ so, let us represent the number ‘1’ as ${{x}^{0}}$
Now, by substituting the above representation of ‘1’ in the required integral then we get,
$\Rightarrow \int{f\left( x \right).dx}=C\int{{{x}^{0}}.dx}$
We know that the standard formula of power rule of integration is given as,
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
By using this power rule to above integral then we get,
$\begin{align}
& \Rightarrow \int{f\left( x \right).dx}=C\left( \dfrac{{{x}^{0+1}}}{0+1}+c \right) \\
& \Rightarrow \int{f\left( x \right).dx}=Cx+Cc \\
& \Rightarrow \int{f\left( x \right).dx}=Cx+{{C}_{1}} \\
\end{align}$
Where, ${{C}_{1}}$ is some other constant not equal to $C$
Therefore, we can conclude that the required integral also called as anti – derivative of constant as,
$\therefore \int{f\left( x \right).dx}=Cx+{{C}_{1}}$
Note: Students may make mistakes mainly by not taking the constant after the integration.
The power rule we used to solve this problem is given as,
$\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$
Here, we can see that constant $'c'$ is very important in case of finding the equation after the integration. Students may miss this constant some times and gives the answer which is not correct.
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