# Evaluate the following integral as limit of sum:

$\int\limits_{0}^{2}{{{e}^{x}}dx}$

Last updated date: 20th Mar 2023

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Answer

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Hint: Apply the formula,

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+......+f\left( a+\left( n-1 \right)h \right) \right]}$, where $h=\dfrac{b-a}{n}.$

The given integral is

$\int\limits_{0}^{2}{{{e}^{x}}dx}$

By definition integral as limit of sum can be written as,

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+....+f\left( a+\left( n-1 \right)h \right) \right]}$, where $h=\dfrac{b-a}{n}$

So, when we are converting an integral to a limit of sum we use the above formula.

Here \[f\left( x \right)\] is the function to be integrated with \[a\] as lower limit and \[b\] as upper limit.

The right hand side is the conversion to limit.

In this question our function is $'{{e}^{x}}'$ with \[0\] as lower limit and \[2\] is upper limit.

So, we get

$\int\limits_{0}^{2}{{{e}^{x}}dx=\left( 2-0 \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{0}}+{{e}^{\left( o+h \right)}}+....+{{e}^{\left( o+\left( n-1 \right)h \right)}} \right]}.......\left( i \right)$

Now here we will find the value of ‘$h$ ‘.

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

Substituting the values of upper and lower limit, we get

$h=\dfrac{2-0}{n}=\dfrac{2}{n}..........\left( ii \right)$

Now substituting the value of the equation$\left( ii \right)$ in equation $\left( i \right)$ , we get

$\int\limits_{0}^{2}{{{e}^{x}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{0}}+{{e}^{\left( o+\dfrac{2}{n} \right)}}+....+{{e}^{\left( o+\left( n-1 \right)\dfrac{2}{n} \right)}} \right]}$

We know ${{e}^{0}}=1$ , so the above equation becomes,

$\int\limits_{0}^{2}{{{e}^{x}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+{{e}^{\left( \dfrac{2}{n} \right)}}+....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}} \right]}...........\left( iii \right)$

Now consider the series,

$1+{{e}^{\left( \dfrac{2}{n} \right)}}+.....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}}.........\left( iv \right)$

This is an geometric progression (GP) series with first term as ${{e}^{\left( \dfrac{2}{n} \right)}}$ , common ratio as ‘${{e}^{\dfrac{2}{n}}}$ ‘

Here we can also observe that there are ‘$n$’ terms in the series.

Now the sum of first $n$ terms of a $GP$ is

${{s}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},r>1$

Where ‘$a$‘ is the first term, ‘$n$ ‘ is the number of terms and ‘$r$ ‘ is the common ratio.

Applying this formula in equation $\left( iv \right)$ series, we get

${{s}_{n}}=\dfrac{1\left( {{e}^{{{\left( \dfrac{2}{n} \right)}^{n}}-1}} \right)}{\left( {{e}^{\dfrac{2}{n}}}-1 \right)}$

${{s}_{n}}=\left( \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right)$

Substituting this value in equation $\left( iii \right)$ we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left( \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right)$

Taking out the constant term, we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{\dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{1}{n}}} \right)$

Now we will multiply and divide by $'2'$ in denominator, we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2\left[ \dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}} \right]} \right).........(iv)$

We know the formula,

$\underset{t\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{t}}-1}{t}=1$

We know, as $n\to \infty $ then $\dfrac{2}{n}\to 0$ .

So, the denominator can be written as,

$\underset{\dfrac{2}{n}\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}}=1$

Substituting this value in the equation (iv), we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2(1)} \right)$

As this is free from $'n'$ term so the value remain same even after applying the limit, so the above equation can be written as,

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\left( \dfrac{1}{2} \right)=({{e}^{2}}-1)$

Hence this is the required integral value.

Note: Student might get confused looking at the series $1+{{e}^{\left( \dfrac{2}{n} \right)}}+.....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}}$.

They might take this as an arithmetic progression series. If we apply the AP series formula, we will get the wrong answer.

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+......+f\left( a+\left( n-1 \right)h \right) \right]}$, where $h=\dfrac{b-a}{n}.$

The given integral is

$\int\limits_{0}^{2}{{{e}^{x}}dx}$

By definition integral as limit of sum can be written as,

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+....+f\left( a+\left( n-1 \right)h \right) \right]}$, where $h=\dfrac{b-a}{n}$

So, when we are converting an integral to a limit of sum we use the above formula.

Here \[f\left( x \right)\] is the function to be integrated with \[a\] as lower limit and \[b\] as upper limit.

The right hand side is the conversion to limit.

In this question our function is $'{{e}^{x}}'$ with \[0\] as lower limit and \[2\] is upper limit.

So, we get

$\int\limits_{0}^{2}{{{e}^{x}}dx=\left( 2-0 \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{0}}+{{e}^{\left( o+h \right)}}+....+{{e}^{\left( o+\left( n-1 \right)h \right)}} \right]}.......\left( i \right)$

Now here we will find the value of ‘$h$ ‘.

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

Substituting the values of upper and lower limit, we get

$h=\dfrac{2-0}{n}=\dfrac{2}{n}..........\left( ii \right)$

Now substituting the value of the equation$\left( ii \right)$ in equation $\left( i \right)$ , we get

$\int\limits_{0}^{2}{{{e}^{x}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{0}}+{{e}^{\left( o+\dfrac{2}{n} \right)}}+....+{{e}^{\left( o+\left( n-1 \right)\dfrac{2}{n} \right)}} \right]}$

We know ${{e}^{0}}=1$ , so the above equation becomes,

$\int\limits_{0}^{2}{{{e}^{x}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+{{e}^{\left( \dfrac{2}{n} \right)}}+....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}} \right]}...........\left( iii \right)$

Now consider the series,

$1+{{e}^{\left( \dfrac{2}{n} \right)}}+.....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}}.........\left( iv \right)$

This is an geometric progression (GP) series with first term as ${{e}^{\left( \dfrac{2}{n} \right)}}$ , common ratio as ‘${{e}^{\dfrac{2}{n}}}$ ‘

Here we can also observe that there are ‘$n$’ terms in the series.

Now the sum of first $n$ terms of a $GP$ is

${{s}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},r>1$

Where ‘$a$‘ is the first term, ‘$n$ ‘ is the number of terms and ‘$r$ ‘ is the common ratio.

Applying this formula in equation $\left( iv \right)$ series, we get

${{s}_{n}}=\dfrac{1\left( {{e}^{{{\left( \dfrac{2}{n} \right)}^{n}}-1}} \right)}{\left( {{e}^{\dfrac{2}{n}}}-1 \right)}$

${{s}_{n}}=\left( \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right)$

Substituting this value in equation $\left( iii \right)$ we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left( \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right)$

Taking out the constant term, we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{\dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{1}{n}}} \right)$

Now we will multiply and divide by $'2'$ in denominator, we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2\left[ \dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}} \right]} \right).........(iv)$

We know the formula,

$\underset{t\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{t}}-1}{t}=1$

We know, as $n\to \infty $ then $\dfrac{2}{n}\to 0$ .

So, the denominator can be written as,

$\underset{\dfrac{2}{n}\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}}=1$

Substituting this value in the equation (iv), we get

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2(1)} \right)$

As this is free from $'n'$ term so the value remain same even after applying the limit, so the above equation can be written as,

$\int\limits_{0}^{2}{{{e}^{x}}}dx=2({{e}^{2}}-1)\left( \dfrac{1}{2} \right)=({{e}^{2}}-1)$

Hence this is the required integral value.

Note: Student might get confused looking at the series $1+{{e}^{\left( \dfrac{2}{n} \right)}}+.....+{{e}^{\dfrac{2}{n}\left( n-1 \right)}}$.

They might take this as an arithmetic progression series. If we apply the AP series formula, we will get the wrong answer.

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