Q. 18 - Calculus

Question

Consider the function 

$F\left(x\right)=\left\{\begin{array}{l} \ln \left|x\right|, if x<0\\ \ln \left|x\right|+4, if x>0\end{array}\right.$

Show that the derivative of this function is the function $f\left(x\right)=\frac{1}{x}$. Compare the graphs of $F\left(x\right)$ and $\ln \left|x\right|$, and discuss how this exercise relates to the second part of Theorem 4.16. 

Step-by-Step Solution

Answer

Proved that $\int f\left(x\right)g\left(x\right)dx\ne \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right) dx\right)$

Step 1. Given information

The given function $F\left(x\right)=\left\{\begin{array}{l} \ln \left|x\right|, if x<0\\ \ln \left|x\right|+4, if x>0\end{array}\right.$

Step 2. Prove &#8747; f ( x ) g ( x ) d x &#8800; &#8747; f ( x ) d x &#8747; g ( x ) &#160; d x

Let 

$$\begin{gathered} f\left(x\right)=x \\ g\left(x\right)=\frac{1}{x} \end{gathered}$$

Now, 

$$\begin{gathered} =\int f\left(x\right)g\left(x\right)dx \\ =\int x\frac{1}{x}dx \\ =\int 1 dx \\ =x+C \end{gathered}$$

And

$$\begin{gathered} =(\int f(x\left)dx\right)(\int g(x\left)dx\right) \\ =(\int x dx) (\int \frac{1}{x}dx) \\ =\left(\frac{x^2}{2}+C\right)\left(\ln x+C\right) \end{gathered}$$

Therefore,$\int f\left(x\right)g\left(x\right)dx\ne \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right) dx\right)$

Hence Proved.