Defining logarithms with integrals: In this chapter we defined the natural logarithm function as the accumulation integral

$\ln x={\int }_0^x \frac{1}{t}dt$

(a) Use the graph of $y=\frac{1}{x}$and this definition to describe the graphical features of $y=\ln x.$

(b) Given this definition of $\ln x,$how would we define the natural exponential function $e^x$? Why is this a better definition for $e^x$than the one we introduced in Definition $1.25$?

Given function$y=\frac{1}{x}$.

The logarithm function is described as $\ln x={\int }_0^x\frac{1}{t}dt$.

The following graph is

The second basic theory of calculus claims that $F^{\text{'}}\left(x\right)=f\left(x\right)$.

This is applied to the definition of logarithm function as localid="1661254900799">lnx=∫0x1tdtas an integration.

Also, the differentiation of$y=\ln x$gives the function $y=\frac{1}{x}$.

Also, the integration of $\ln x={\int }_0^x\frac{1}{t}dt$gives the function $y=\ln x$.

The graph of $y=\ln x$is only valid in the localid="1661255098361">x>0.

Consequently, the space beneath the curve oflocalid="1661255101566">y=1xand thex-axis is equal to localid="1661255118403">y=lnx.

The given function is$y=\frac{1}{x}$.

The graph is

Using the second fundamental calculus theory,

$F^{\text{'}}\left(x\right)=f\left(x\right)$

And,

$\frac{d{e}^x}{dx}=e^x$

Also the differentiation of $y=e^x$gives the function $\frac{d{e}^x}{dx}=e^x$.

Also the integration of $e^x={\int }_0^x{e}^{t}dt$gives the function $y=e^x$.

The graph of $y=e^x$is only valid in the $y>0$.

Consequently, the space beneath the curve of $y=e^x$ and the x-axis is equal to $y=e^x$.

This is a good estimate of the area under the curve. Since it also adheres to the second fundamental calculus theory, it is a superior definition. $F^{\text{'}}\left(x\right)=f\left(x\right)$.

As a result, the exponential function's definition islocalid="1661254990037">ex=∫0xetdt.