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

Obtain the value of ∫01exdx\int_0^1 {{e^x}dx} as the limit of a sum.

Explanation

Solution

Hint: The formula for integral of ∫abf(x)dx\int_a^b {f(x)dx} in terms of limit of a sum is lim⁡n→∞1n(f(a)+f(a+h)+....f(a+(n−1)h))\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(f(a) + f(a + h) + ....f(a + (n - 1)h)) where h=b−anh = \dfrac{{b - a}}{n}. Use this to evaluate ∫01exdx\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:
∫abf(x)dx=lim⁡n→∞1n(f(a)+f(a+h)+....+f(a+(n−1)h)).......(1)\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=b−an............(2)h = \dfrac{{b - a}}{n}............(2)
The value of h from the given integral is given as follows:
h=1−0nh = \dfrac{{1 - 0}}{n}
h=1n...........(3)h = \dfrac{1}{n}...........(3)
From equation (1), we observe that the function f(x) is ex{e^x} and a is 0 and b is 1. Then, we have:
f(0)=e0=1f(0) = {e^0} = 1
f(h)=ehf(h) = {e^h}
.
.
f((n−1)h)=e(n−1)hf((n - 1)h) = {e^{(n - 1)h}}
Substituting the above values in equation (1), we have:
∫01exdx=lim⁡n→∞1n(f(0)+f(h)+....+f((n−1)h))\int_0^1 {{e^x}dx = } \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}(f(0) + f(h) + .... + f((n - 1)h))
∫01exdx=lim⁡n→∞1n(1+eh+e2h....+e(n−1)h)........(3)\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+eh+e2h....+e(n−1)h1 + {e^h} + {e^{2h}}.... + {e^{(n - 1)h}} are the sum of G.P. with the first term as 1 and common ratio as eh{e^h} and has n terms. The sum is given as follows:
a+ar+ar2+....+arn−1=a(rn−1)r−1a + ar + a{r^2} + .... + a{r^{n - 1}} = \dfrac{{a({r^n} - 1)}}{{r - 1}}
1+eh+e2h....+e(n−1)h=enh−1eh−1............(4)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:
∫01exdx=lim⁡n→∞1nenh−1eh−1\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→∞⇒h→0n \to \infty \Rightarrow h \to 0
Hence, we have the following expression.
∫01exdx=lim⁡h→0e1h×h−1eh−1h\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 lim⁡x→0ex−1x\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} is equal to 1, hence, we have:
∫01exdx=lim⁡h→0e1−11\int_0^1 {{e^x}dx = } \dfrac{{\mathop {\lim }\limits_{h \to 0} {e^1} - 1}}{1}
∫01exdx= e−1\int_0^1 {{e^x}dx = } {\text{ }}e - 1
Hence, the value of the given integral is e−1e - 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.