Question

How do you find the indefinite integral of $\int {12{x^2}dx?}$

Answer 1

To find the indefinite integral of the given function, first consider the given integration to be some variable (say “I”) then use property of indefinite integral and separate the constant part from the integrand or the function, then integrate the integrand and then finally write the integral with the constant part multiplied with it.

Complete step by step answer:
In order to find the indefinite integral of $\int {12{x^2}dx}$, we will first consider the given integration to be $I$, and then separate the constant from integral part and then we will integrate the rest function as follows
$I = \int {12{x^2}dx}$
Multiplying and dividing $12$ both sides, we will get
$\dfrac{{12}}{{12}} \times I = \dfrac{{12}}{{12}} \times \int {12{x^2}dx}$
We can further write it as

\Rightarrow I = 12\int {{x^2}dx} \\\ $$ Now, doing the integration of the integrand $f(x) = {x^2}$ with help of the integration formula $\int {{x^n}dx} = \dfrac{{{x^{n - 1}}}}{{n - 1}} + c$ Using the above integration formula, to integrate the given function, we will get $$I = 12\int {{x^2}dx} \\\ \Rightarrow I = 12\left( {\dfrac{{{x^{2 + 1}}}}{{2 + 1}} + c} \right) \\\ \Rightarrow I = 12\left( {\dfrac{{{x^3}}}{3} + c} \right) \\\ \Rightarrow I = 4{x^3} + 12c \\\ $$ We know that “c” is an arbitrary constant, so we can express $12c$ as another variable as follows $$ \therefore I = 4{x^3} + C$$ **Therefore $$4{x^3} + C$$ is the required integration of the indefinite integral $$\int {12{x^2}dx} $$.** **Note:** After integration of the indefinite integrals always write $ \pm c$ in its integration part, this $ \pm c$ is an arbitrary constant or the constant of integration. Let us see the necessity of writing $ \pm c$ at the integration part with an example, derivative of ${x^2} + 23$ is $2x$ and the derivative of ${x^2} - 2$ is also $2x$, so in order to maintain the constant in the integration part, we write $ \pm c$. Also $c$ is an arbitrary constant that’s why when we multiplied it with $12$, we have written it $C$.