Question

(a) What is an antiderivative of a function f?

(b) Suppose \({{\bf{F}}_{\bf{1}}}\) and \({{\bf{F}}_2}\) are both antiderivatives of on an interval I. How are \({{\bf{F}}_{\bf{1}}}\) and \({{\bf{F}}_2}\) related?

Short answer

(a) The antiderivative of a function f is F, where \(\int {f\left( x \right)} dx = F\left( x \right) + C\).

(b) The relation between \({F_1}\) and \({F_2}\) is given as \({F_1} = {F_2} + c\).

Step-by-step solution

Part (a)

The antiderivative of function \(f\left( x \right)\) is a function whose output or we can say derivative is equal to \(F'\left( x \right)\) for all x in the domain of the function \(f\left( x \right)\). That is, if \(F'\left( x \right) = f\left( x \right)\), then we can say \(F\left( x \right)\) is the antiderivative of \(f\left( x \right)\).

Where, \(\int {f\left( x \right)} dx = F\left( x \right) + C\), for any constant C.

Example

Consider two functions as below:

\(F\left( x \right) = {x^3} - 5x + 2\)and \(f\left( x \right) = 3{x^2} - 5\)

Check whether \(f\left( x \right)\) is the antiderivative of \(F\left( x \right)\).

Find derivative of \(F\left( x \right)\).

\(F'\left( x \right) = 3{x^2} - 5\)

As, \(F'\left( x \right) = f\left( x \right)\), so \(f\left( x \right)\) is the antiderivative of \(F\).

Eventually, \(F\) is an antiderivative of \(f\).

Part (b)

Two given functions are \({F_1}\) and \({F_2}\), which are derivative antiderivative of \(f\) on any \(I\). Then,

\(\int f = {F_1}{\rm{ or }}f = \frac{{d{F_1}}}{{dx}}{\rm{ }}......\left( 1 \right)\)

\(\int f = {F_2}{\rm{ or }}f = \frac{{d{F_2}}}{{dx}}{\rm{ }}......\left( 2 \right)\)

Subtract equation (2) from (1).

\(\begin{aligned}{c}\frac{{d{F_1}}}{{dx}} - \frac{{d{F_2}}}{{dx}} = f - f\\\frac{d}{{dx}}\left( {{F_1} - {F_2}} \right) = 0\end{aligned}\)

Now, integrate both the sides.

\(\begin{aligned}{c}{F_1} - {F_2} = c\\{F_1} = {F_2} + c\end{aligned}\)

Where, c is a constant.

So, a relation between \({F_1}\) and \({F_2}\) is \({F_1} = {F_2} + c\).