Question

What is the integral of a constant?

Answer 1

Hint: In this type of question we have to use the concepts of integration and some rules of indices. We know that anything raised to zero is always equal to 1 that is \[{{x}^{0}}=1\]. Also we know that \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}\] where \[c\] is an arbitrary constant which is also known as constant of integration

Complete step-by-step answer:
Now we have to find the integral of a constant say k.
For this let us consider,
\[= \int{kdx}\]
Now as k is a constant we can rewrite the integral as
\[= k\int{1dx}\]
By the rule of indices we can write \[{{x}^{0}}=1\] and hence
\[= k\int{{{x}^{0}}dx}\]
By using the rule, \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}\] we get,
\[= k\left( \dfrac{{{x}^{0+1}}}{0+1} \right)+c\] where \[c\] is an arbitrary constant which is also known as constant of integration
\[= k\left( \dfrac{{{x}^{1}}}{1} \right)+c\]
On simplifying we can write,
\[= kx+c\]
Hence, we can say that the integral of a constant say k is given by, \[kx+c\]
In other words, we can write, \[\int{kdx=kx+c}\]

Note: In this type of question one of the students may use another way to find the integral of a constant in following manner:
We know that by the rules of differentiation \[\dfrac{d}{dx}\left( kx+c \right)=k\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( c \right)=k\] where \[k\] and \[c\] are constant and we know that the derivative of a constant is zero.
\[= \dfrac{d}{dx}\left( kx+c \right)=k\]
Now taking integral of both sides we get,
\[= \int{\left[ \dfrac{d}{dx}\left( kx+c \right) \right]dx}=\int{kdx}\]
As we know that integration and differentiation are inverse of each other,
\[= kx+c=\int{kdx}\]
Hence, we can say that the integral of a constant say k that is \[\int{kdx}\] equals to \[kx+c\] where \[c\] is an arbitrary constant which is also known as constant of integration.