Answer
Verified
411.3k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE