Q. 91 - Calculus

Question

Consider the following formula for anti differentiating power functions: If $$\begin{gathered} f\text{'}\left(x\right)=x^k, \\ k\ne -1 \end{gathered}$$, then $f\left(x\right)=\frac{1}{k+1}{x}^{k+1}+C$ for some constant C.

Part (a): Prove this antidifferentiation formula. You may assume that any two functions with the same derivative differ by a constant, as we will prove in Section 3.2.

Part (b): What part of your argument from part (a) breaks down when $f\left(x\right)=x^{-1}$?

Step-by-Step Solution

Answer

Part (a): To prove the anti differentiation formula, assume two functions and differentiate the following.

In both cases the derivative of f is same, but the functions are not same. They differ by a constant.

Part (b): The coefficient of the variable of the function has problem while taking anti derivative. 

Part (a) Step 1. Given information.

Consider the given question,

$$\begin{gathered} f\text{'}\left(x\right)=x^k, \\ k\ne -1 \end{gathered}$$

Then, $f\left(x\right)=\frac{1}{k+1}{x}^{k+1}+C$

Part (a) Step 2. Prove the anti differentiating power formula.

Differentiate with respect to x,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left[\frac{1}{k+1}{x}^{k+1}+20\right] \\ =\frac{1}{k+1}{\left(k+1\right)}^{k+1-1} \\ =x^k \end{gathered}$$

For another function $f\left(x\right)=\frac{1}{k+1}{x}^{k+1}+100$,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left[\frac{1}{k+1}{x}^{k+1}+100\right] \\ =\frac{1}{k+1}\left(k+1\right)x^{k+1-1} \\ =x^k \end{gathered}$$

In both cases the derivative of f is same, but the functions are not same. They differ by a constant.

Part (b) Step 1. Take the anti derivative for the function f x = x - 1 .

Consider the function $f\left(x\right)=x^{-1}$, if the anti derivative is taken then,

$\frac{1}{-1+1}{x}^{-1+1}=\frac{1}{0}\left(x^0\right)$

This is undefined.

Therefore, the coefficient of the variable of the function has problem while taking anti derivative.