QUESTION

# Obtain the value of $\int_0^1 {{e^x}dx}$ as the limit of a sum.

Hint: The formula for integral of $\int_a^b {f(x)dx}$ in terms of limit of a sum is $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(f(a) + f(a + h) + ....f(a + (n - 1)h))$ where $h = \dfrac{{b - a}}{n}$. Use this to evaluate $\int_0^1 {{e^x}dx}$ as the limit of a sum.

We know the formula for limit of a sum as follows:
$\int_a^b {f(x)dx = } \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(f(a) + f(a + h) + .... + f(a + (n - 1)h)).......(1)$
The value of h is given as below:
$h = \dfrac{{b - a}}{n}............(2)$
The value of h from the given integral is given as follows:
$h = \dfrac{{1 - 0}}{n}$
$h = \dfrac{1}{n}...........(3)$
From equation (1), we observe that the function f(x) is ${e^x}$ and a is 0 and b is 1. Then, we have:
$f(0) = {e^0} = 1$
$f(h) = {e^h}$
.
.
$f((n - 1)h) = {e^{(n - 1)h}}$
Substituting the above values in equation (1), we have:
$\int_0^1 {{e^x}dx = } \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(f(0) + f(h) + .... + f((n - 1)h))$
$\int_0^1 {{e^x}dx = } \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(1 + {e^h} + {e^{2h}}.... + {e^{(n - 1)h}})........(3)$
The terms $1 + {e^h} + {e^{2h}}.... + {e^{(n - 1)h}}$ are the sum of G.P. with the first term as 1 and common ratio as ${e^h}$ and has n terms. The sum is given as follows:
$a + ar + a{r^2} + .... + a{r^{n - 1}} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
$1 + {e^h} + {e^{2h}}.... + {e^{(n - 1)h}} = \dfrac{{{e^{nh}} - 1}}{{{e^h} - 1}}............(4)$
Substituting equation (4) in equation (3), we get:
$\int_0^1 {{e^x}dx = } \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\dfrac{{{e^{nh}} - 1}}{{{e^h} - 1}}$
Using equation (2), we change the limits in terms of h.
$n \to \infty \Rightarrow h \to 0$
Hence, we have the following expression.
$\int_0^1 {{e^x}dx = } \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{\dfrac{1}{h} \times h}} - 1}}{{\dfrac{{{e^h} - 1}}{h}}}$
We know that the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x}$ is equal to 1, hence, we have:
$\int_0^1 {{e^x}dx = } \dfrac{{\mathop {\lim }\limits_{h \to 0} {e^1} - 1}}{1}$
$\int_0^1 {{e^x}dx = } {\text{ }}e - 1$
Hence, the value of the given integral is $e - 1$.

Note: You can cross-check your answer by actually evaluating the integral by using the normal integration method to find the answer. Note that it is only for cross-checking.