Question

find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antiderivative.)

$f(x)=x(18x + 4)$

$f(x)=$

Explanation:

Step1: Expand the function

First, expand \(f(x)=x(18x + 4)=18x^{2}+4x\).

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For the term \(18x^{2}\), \(\int18x^{2}dx=18\int x^{2}dx\). Applying the power rule, \(18\times\frac{x^{2+1}}{2 + 1}=18\times\frac{x^{3}}{3}=6x^{3}\).
For the term \(4x\), \(\int4xdx=4\int xdx\). Applying the power rule, \(4\times\frac{x^{1+1}}{1+1}=4\times\frac{x^{2}}{2}=2x^{2}\).

Step3: Add the constant of integration

The most general antiderivative \(F(x)\) of \(f(x)\) is \(F(x)=6x^{3}+2x^{2}+C\).

Step4: Check by differentiation

Differentiate \(F(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(F^\prime(x)=(6x^{3}+2x^{2}+C)^\prime\).
\((6x^{3})^\prime=6\times3x^{2}=18x^{2}\), \((2x^{2})^\prime=2\times2x = 4x\), and \((C)^\prime=0\).
So \(F^\prime(x)=18x^{2}+4x=f(x)\).

Answer:

\(F(x)=6x^{3}+2x^{2}+C\)