Question

Write an antiderivative for each of the following functions using the method of Inspection.
(i) $\cos 2x$
(ii) $3{x^2} + 4{x^3}$
(iii) $\dfrac{1}{x},x \ne 0$

Answer 1

Here, we have to find the antiderivative for the given functions. Anti derivative is also termed as an integration which is inverse of differentiation. We will use the differentiation formula to find the antiderivative for the given function. The anti derivative of a function is defined as a function whose derivative is equal to the given function.

Formula Used:
We will use the following differentiation formula:
1.$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
2.$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
3.$\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$

Complete step-by-step answer:
(i) $\cos 2x$
Let $f\left( x \right) = \cos 2x$
By using the differentiation formula $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
We know that ${\left( {\sin x} \right)^{'}} = \cos x$
By replacing $x$ by $2x$, we get
$\Rightarrow {\left( {\sin 2x} \right)^{'}} = 2\cos 2x$
By rewriting the equation, we get
$\Rightarrow \dfrac{1}{2} \times {\left( {\sin 2x} \right)^{'}} = 2\cos 2x$
Anti derivative of $\left( {\cos 2x} \right)$ is $\dfrac{1}{2} \times \sin 2x$ .
(ii) $3{x^2} + 4{x^3}$
Let $f\left( x \right) = 3{x^2} + 4{x^3}$
By using the differentiation formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
We know that ${\left( {{x^3}} \right)^{'}} = 3{x^{3 - 1}}$
$\Rightarrow {\left( {{x^3}} \right)^{'}} = 3{x^2}$……………………………………………………………….$\left( 1 \right)$

By using the differentiation formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
We know that ${\left( {{x^4}} \right)^{'}} = 4{x^{4 - 1}}$
$\Rightarrow {\left( {{x^4}} \right)^{'}} = 4{x^3}$……………………………………………………………….$\left( 2 \right)$
By adding the equation, we get
$\Rightarrow {\left( {{x^3} + {x^4}} \right)^{'}} = 3{x^2} + 4{x^3}$
Anti derivative of $3{x^2} + 4{x^3}$ is ${x^3} + {x^4}$ .
(iii) $\dfrac{1}{x},x \ne 0$
Let $f\left( x \right) = \dfrac{1}{x},x \ne 0$
By using the differentiation formula $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
We know that ${\left( {\log x} \right)^{'}} = \dfrac{1}{x}$
Anti derivative of $\dfrac{1}{x},x \ne 0$ is $\log x$
Therefore,
Anti derivative of $\left( {\cos 2x} \right)$ is $\dfrac{1}{2} \times \sin 2x$ .
Anti derivative of $3{x^2} + 4{x^3}$ is ${x^3} + {x^4}$ .
Anti derivative of $\dfrac{1}{x},x \ne 0$ is $\log x$ .

Note: We know that method of Inspection is a method of guessing the solutions for the given equation. Anti derivative is found by using the concept of Integration. Whenever we are finding the antiderivative, the anti derivative of a function is always accompanied with an arbitrary constant. But we neglect the arbitrary constant since it is not necessary to include an arbitrary constant of the indefinite Integral in the computation of the definite integral. Integration is a method of finding the whole whereas differentiation is a method of finding a part from the whole.