
Solve and find the value of
$ \int{3xdx} $
Answer
580.2k+ views
Hint: We are given the function 3x and we have to find it integral. We can use the property of the integral of constant times a function to simplify the integration and then use the formula for the integration of x times dx, which is $ \int{xdx}=\dfrac{{{x}^{2}}}{2} $ , to obtain the answer to this given question.
Complete step-by-step answer:
We know that the integration of a constant c times a function f is given by
$ \int{cf(x)dx=c\int{f(x)dx..................(1.1)}} $
i.e. the integral of constant times a function is that constant multiplied by integral of the function.
Now, taking f(x)=x and c=3 in equation (1.1), we obtain
$ \int{3xdx=3\int{xdx..................(1.2)}} $
Now, we also know that the indefinite integral of x is given by
$ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}..........................(1.3)} $
Taking n=1 in (1.3), we have
$ \int{xdx}=\int{{{x}^{1}}dx=\dfrac{{{x}^{1+1}}}{1+1}=\dfrac{{{x}^{2}}}{2}..............(1.4)} $
Thus, using the value obtained in equation (1.4) in equation (1.2), we obtain
\[\begin{align}
& \int{3xdx=3\int{xdx=3\times \dfrac{{{x}^{2}}}{2}}} \\
& \Rightarrow \int{3xdx=\dfrac{3{{x}^{2}}}{2}} \\
\end{align}\]
Which is the required answer to this question.
Note: We could also have solved this problem using change of variables. If we had taken $ 3x=s $ , then we should have $ dx=\dfrac{ds}{3} $ and using (1.4), we could have written the integral as $ \begin{align}
& \int{3xdx=\int{s\dfrac{ds}{3}=\dfrac{1}{3}\int{sds}}} \\
& =\dfrac{1}{3}\times \dfrac{{{s}^{2}}}{2}=\dfrac{3x\times 3x}{6}=\dfrac{3{{x}^{2}}}{2} \\
\end{align} $
Which is the same answer as we had obtained in the solution. However, while using change of variables, we should be careful to express dx in terms of the new variable and then perform the integral.
Complete step-by-step answer:
We know that the integration of a constant c times a function f is given by
$ \int{cf(x)dx=c\int{f(x)dx..................(1.1)}} $
i.e. the integral of constant times a function is that constant multiplied by integral of the function.
Now, taking f(x)=x and c=3 in equation (1.1), we obtain
$ \int{3xdx=3\int{xdx..................(1.2)}} $
Now, we also know that the indefinite integral of x is given by
$ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}..........................(1.3)} $
Taking n=1 in (1.3), we have
$ \int{xdx}=\int{{{x}^{1}}dx=\dfrac{{{x}^{1+1}}}{1+1}=\dfrac{{{x}^{2}}}{2}..............(1.4)} $
Thus, using the value obtained in equation (1.4) in equation (1.2), we obtain
\[\begin{align}
& \int{3xdx=3\int{xdx=3\times \dfrac{{{x}^{2}}}{2}}} \\
& \Rightarrow \int{3xdx=\dfrac{3{{x}^{2}}}{2}} \\
\end{align}\]
Which is the required answer to this question.
Note: We could also have solved this problem using change of variables. If we had taken $ 3x=s $ , then we should have $ dx=\dfrac{ds}{3} $ and using (1.4), we could have written the integral as $ \begin{align}
& \int{3xdx=\int{s\dfrac{ds}{3}=\dfrac{1}{3}\int{sds}}} \\
& =\dfrac{1}{3}\times \dfrac{{{s}^{2}}}{2}=\dfrac{3x\times 3x}{6}=\dfrac{3{{x}^{2}}}{2} \\
\end{align} $
Which is the same answer as we had obtained in the solution. However, while using change of variables, we should be careful to express dx in terms of the new variable and then perform the integral.
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