# What is the area of the line 2^x using linear approximation.

## Question:

What is the area of the line {eq}2^x {/eq} using linear approximation.

## Linearization and Area:

Let us suppose that any function {eq}y=f(x) {/eq} of one variable is given.

Then the Linearization of {eq}y=f(x) {/eq} with respect to any point {eq}x=a {/eq} is given by:

{eq}L(x) = f(a) + \left( {x - a} \right)f'(a) {/eq}

In this case, the given function is:{eq}f(x) ={2^x}, \; a = 0. {/eq}

Here:

{eq}\displaystyle f(x) ={2^x}\Rightarrow f\left(0\right) = 1 \\ \displaystyle f'(x) = {2^x}\ln 2 \Rightarrow f'\left(0\right) =\ln 2 \\ {/eq}

Hence the Linearization of the function is given by:

{eq}\eqalign{ L(x)& = f\left( 0 \right) + \left( {x - 0} \right)f'\left( 0 \right) \cr & = 1 + \left( {x - 0} \right)\left( {\ln 2} \right) \cr & = 1 + x\left( {\ln 2} \right) \cr} {/eq}

Hence the required intercepts of the line: {eq}y = 1 + x\left( {\ln 2} \right) {/eq} are given as {eq}(0,1) {/eq} and {eq}\displaystyle (-\frac{1}{\left( {\ln 2} \right) },0) {/eq}

Hence the required area is: {eq}\displaystyle A=\frac{1}{2}ab {/eq}

Which means:

{eq}\displaystyle A=\frac{1}{\left( {2 \ln 2} \right) } {/eq} 