Question

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

Answer 1

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.

Complete step-by-step answer:
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.