# Use definition 2 to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

$f ( x ) = \dfrac { 2 x } { x ^ { 2 } + 1 } , 1 \leq x \leq 3$

This article aims to write the expression for the area under the graph. The article uses the concept of definition $2$ to find the expression for the area under the graph. The definition $2$ states that:

$Area =\lim_{ n \to \infty } \Delta x \sum_{ i = 1 } ^ { n } f( x _ { i } )$

Where:

$\Delta = \dfrac { b – a } { n }$

The definition $2$ states that:

$Area =\lim_{ n \to \infty } \Delta x \sum_{i=1}^{n} f(x_{i})$

Where:

$\Delta = \dfrac { b – a } { n }$

If we choose $x_{i}$ as the right endpoint of each interval, then:

$Area =\lim_{ b \to \infty } \Delta x \sum_{ i = 1 } ^ { n } f( a + i \Delta x )$

$f ( x ) = \dfrac { 2 x } { x ^ { 2 } + 1 }$

$a = 1, b = 3$

Hence,

$\Delta x = \dfrac { b – a } { n } = \dfrac { 3 – 1 } { n } = \dfrac { 2 } { n }$

$Area =\lim_{ b \to \infty } \Delta x \sum_{ i = 1 } ^ { n } f ( a + i \Delta x ) = \lim_{ n \to \infty } \dfrac { 2 } { n } \sum_{ i = 1 } ^ { n } f ( 1 + i . \dfrac { 2 } { n } )$

$= \lim_{n \to \infty} \dfrac{2}{n} \sum_{i=1}^{n} \dfrac{ 2[1 + \dfrac { 2i } { n } ] }{[1 + \dfrac { 2 i } { n }] ^ { 2 } + 1 }$

The expression for the area under the curve is $\lim_{n \to \infty} \dfrac{2}{n} \sum_{i=1}^{n} \dfrac{2[1+\dfrac{2i}{n}]}{[1+\dfrac{2i}{n}]^{2}+1}$.

## Numerical Results

The expression for the area under the curve is $\lim_{n \to \infty} \dfrac{2}{n} \sum_{i=1}^{n} \dfrac{2[1+\dfrac{2i}{n}]}{[1+\dfrac{2i}{n}]^{2}+1}$.

## Example

Use definition $2$ to find an expression for area under the graph and with the limit. Do not evaluate the limit.

$f ( x ) = \dfrac { 4 x } { x ^ { 2 } – 1 } , 1 \leq x \leq 4$

Solution

The definition $2$ states that:

$Area =\lim_{ n \to \infty } \Delta x \sum_{i=1}^{n} f(x_{i})$

Where:

$\Delta = \dfrac{b-a}{n}$

If we choose $x_{i}$ as the right endpoint of each interval, then:

$Area =\lim_{ b \to \infty } \Delta x \sum_{i=1}^{n} f(a+i\Delta x )$

$f(x) = \dfrac{4x}{x^{2}-1}$

$a = 1, b = 4$

Hence,

$\Delta x = \dfrac{b-a}{n} = \dfrac{4-1}{n} = \dfrac{3}{n}$

$Area =\lim_{ b \to \infty } \Delta x \sum_{i=1}^{n} f(a+i\Delta x ) = \lim_{n \to \infty} \dfrac{3}{n} \sum_{i=1}^{n} f(1+i.\dfrac{3}{n})$

$= \lim_{n \to \infty} \dfrac{3}{n} \sum_{i=1}^{n} \dfrac{4[1+\dfrac{3i}{n}]}{[1+\dfrac{3i}{n}]^{2}-1}$

The expression for the area under curve is $\lim_{n \to \infty} \dfrac{3}{n} \sum_{i=1}^{n} \dfrac{4[1+\dfrac{3i}{n}]}{[1+\dfrac{3i}{n}]^{2}-1}$.