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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:

\[ \Delta = \dfrac { b – a } { n } \]

Expert Answer

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:

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

In this article:

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

\[a = 1, b = 3\]

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:

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

In this article:

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

\[a = 1, b = 4\]

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} $.

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