Question

What is the Integration of $dx$ ?

Answer 1

Hint : In order to determine the integration of $dx$ write it as $1dx$ which we can further write as ${x^0}dx$ . Now, Integrate the values obtained using the power rule that is $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1}$ and hence we get our result. Always add constant terms at last after integrating the equation.

Complete step by step solution:
We are given with the value $dx$ .
First write the value of $dx$ as $1dx$ .Since, we know that ${x^0} = 1$ , so write the same in the previous value and we get:
$dx = 1dx = {x^0}dx$
Now, Integrate the value and we get:
$\int {dx} = \int {1dx} = \int {{x^0}dx}$
Using the power rule which is $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1}$ , we can solve the integration value further and hence, we get:
$\int {{x^0}dx = \dfrac{{{x^{0 + 1}}}}{{0 + 1}}} = \dfrac{{{x^1}}}{1} = x$
But this cannot be the accurate value because when we derive an equation its constant is removed, so let’s add $C$ as constant value and we get:
$\int {{x^0}dx} = x + C$
Hence, the Integration of $dx$ is $x + C$
So, the correct answer is “$x + C$”.

Note : It’s very important to add a constant after integrating the equation because when we derive an equation their constant is removed. For example, let’s see an equation: $2x + 2$ Derivative this with respect to $x$ and we get: $\dfrac{{d(2x + 2)}}{{dx}} = \dfrac{{d2x}}{{dx}} + \dfrac{{d2}}{{dx}} = 2 + 0 = 2$ .Because we know that derivation of constant term is zero. Now Let’s check and integrate the answer obtained and we get:
$\int {2dx} = 2x$ and we can see that the constant term is lost. That’s why we need to add a constant term at last after integration.