Questions & Answers

Question

Answers

Answer
Verified

Definite integrals are used when the limit is defined, to generate, a unique value. Indefinite integrals are implemented when the limits of the integrand are not specified. Definite integral is denoted as

\[\int\limits_a^b {f(x)\,dx} \]

Where \[f(x)\] is a function of x or integrand and \['dx'\] is the integrating agent. The equation indicates the integral of \[f(x)\] with respect to x. The definite integral of any function can be expressed as the limit of sum. Now to solve integration firstly integrated the expression then put the limits. Formula which is used for solving the integral as the limit if sum is

\[\int\limits_a^b {f(x)} \,dx = (b - a)\mathop {\lim }\limits_{n \to } \dfrac{1}{n}\left[ {f(a) + f(a - h) + .... + .... + f(a + (n - 1)h} \right]\]

Where \[h = \dfrac{{b - a}}{n}\]

\[a = \]lower limit, \[b = \]upper limit

\[\int\limits_o^2 {({x^2} + 1} )\]

Here \[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {n + \dfrac{2}{3}(1 - \dfrac{1}{n})(2 - \dfrac{1}{n})} \right\}\]

\[h - \dfrac{{b - a}}{n} = \dfrac{{2 - 0}}{n} = \dfrac{2}{n}\]

Therefore,

Solving square of each

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {1 + \dfrac{4}{{{n^2}}} + 1 + \dfrac{{16}}{{{n^2}}} + 1 + \dfrac{{{{(2n - 2)}^2}}}{{{n^2}}} + 1} \right]\]

Can also be written as

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {\left[ {1 + 1 + 1 + 1..... + (ntimes)} \right] + \dfrac{1}{{{n^2}}}\left[ {{2^2} + {4^2}.....{{(2n - 2)}^2}} \right]} \right\}\]

Taking n common

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {\left[ n \right] + \dfrac{1}{{{n^2}}}{2^2}\left[ {{1^2} + {2^2}.....{{(n - 1)}^2}} \right]} \right\}\]

Solving

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {n + \dfrac{4}{{{n^2}}}\left[ {\dfrac{{(n - 1)n(2n - 1)}}{6}} \right]} \right\}\]

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {n + \dfrac{2}{3}\dfrac{{(n - 1)(2n - 1)}}{n}} \right\}\]

4 and 6 has common factor 2 so cancelled and written as

\[ = 2\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left\{ {n + \dfrac{2}{3}(1 - \dfrac{1}{n})(2 - \dfrac{1}{n})} \right\}\]

As

\[n \to \infty ,\,\dfrac{1}{n} \to 0\]

Therefore we have

\[\int_0^2 {\left( {{x^2} + 1} \right)} dx = 2\left[ {1 + \dfrac{4}{3}} \right] = \dfrac{{14}}{3}\]

Now, \[f(0) = {0^2} + 1 = 1\]

i.e. \[f(0) = 1\]

Hence, \[f\left( {0\, - \,\dfrac{2}{n}} \right) = {\left( {\dfrac{{ - 2}}{n}} \right)^2} + 1 = \dfrac{4}{{{n^2}}} + 1\].

2. When \[1\] is added n times its sum is ānā.

3. Sum of \[{1^2} + {2^2} + .......{(n - 1)^2} = \dfrac{{n(n - 1)(2n - 1)}}{6}\]