Question

Express $\int_a^b {{e^x}dx}$ as a limit of sum and evaluate it.

Answer 1

We will first write the integral in the form of limit of sum as $\int_a^b {f\left( x \right)dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( a \right) + f\left( {a + h} \right) + ..... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$, where $h = \dfrac{{b - a}}{n}$ and $n$ represents the number of term. Then, simplify the expression and apply the limits to get the required answer.

Complete step-by-step answer:
When we have to write a function as a limit of sum, the integral is written as
$\int_a^b {f\left( x \right)dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( a \right) + f\left( {a + h} \right) + ..... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$, where $h = \dfrac{{b - a}}{n}$ and $n$ represents the number of term.
In the given expression,
$f\left( x \right) = {e^x}$
Then,
$\int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {{e^a} + {e^{a + h}} + ..... + {e^{a + \left( {n - 1} \right)h}}} \right]$
We can rewrite the expression as
$\int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {{e^a} + {e^a}{e^h} + ..... + {e^a}{e^{\left( {n - 1} \right)h}}} \right]$
The expression inside the bracket represents G.P., where first term is ${e^a}$ and the common ratio is ${e^h}$ and there are $n$ terms.
Now, the sum of G.P. is given by $\dfrac{{A\left( {{R^n} - 1} \right)}}{{R - 1}}$, where $A$ is the first term of the GP and $R$ is the common ration of G.P.
Therefore, the sum is $\dfrac{{{e^a}\left( {{e^{hn}} - 1} \right)}}{{{e^h} - 1}}$
On substituting the value, we will get,
$\int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {\dfrac{{{e^a}\left( {{e^{hn}} - 1} \right)}}{{{e^h} - 1}}} \right]$
Which is also equivalent to $\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{{e^a}\left( {{e^{hn}} - 1} \right)}}{{\dfrac{{{e^h} - 1}}{h}}}} \right]$
It is known that $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^x} - 1}}{x}} \right) = 1$
Then, $\int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} {e^a}\left( {{e^{hn}} - 1} \right)$
Also, $h = \dfrac{{b - a}}{n}$, which will give
$\int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} {e^a}\left( {{e^{\left( {\dfrac{{b - a}}{n}} \right)n}} - 1} \right) \\\ \Rightarrow \int_a^b {{e^x}dx} = \mathop {\lim }\limits_{h \to 0} {e^a}\left( {{e^{b - a}} - 1} \right) \\\$
Now, apply the limit and open the bracket,
$\int_a^b {{e^x}dx} = {e^a}{e^{b - a}} - {e^a} \\\ \Rightarrow \int_a^b {{e^x}dx} = {e^b} - {e^a} \\\$
Therefore, the value of $\int_a^b {{e^x}dx}$ is ${e^b} - {e^a}$

Note: We cannot apply the values of limit when the given expression will give indeterminate form. Also, the value of $e$ is greater than 1, hence, we have used the formula $\dfrac{{A\left( {{R^n} - 1} \right)}}{{R - 1}}$ for the sum of G.P. We have simplified the expression using the identity ${a^{m + n}} = {a^m}{a^n}$.