Question

Question:-

$\rm \: find \: \int_{0}^{2} ( {x}^{2} + 1)dx \: as \: the \: limit \: of \: a \: sum$
Note :- Solve by using limit of a sum method ​

Answer 1

Answer

To find integral using limit of sum method :-

$\displaystyle\boxed{\sf \int_{a}^{b} f( {x})dx = (b - a)\lim_{n \to \infty}( \frac{1}{n} )[f(a) + f(a + n) + f(a + \{n - 1\}h)]}$

where $\sf h = \frac{b-a}{n}$

For given question :-

• $\rm a = 0$
• $\rm b = 2$
• $\rm f(x) = ( x^2 + 1 )$
• $\rm h = \frac{2}{n}$

$\displaystyle\sf \int_{0}^{2} ( {x}^{2} + 1)dx = 2 \lim_{n \to \infty}( \frac{1}{n} )[f(0) + f(\frac{2}{n}) + f(\frac{4}{n}) + ...... + f(\frac{2 \{{n - 1} \}}{n} )]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[1 + \{ \frac{2^2}{n^2} + 1\} + \{ \frac{4^2}{n^2} + 1 \} + ..... + \{ \frac{(2n-2)^2}{n^2} + 1 \}]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )([ 1 + 1 + 1 ....... 1(n - times) ] + \frac{1}{n^2} )[2^2 + 4^2 +... (2n-2)^2]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[ n + \frac{2^2}{n^2}(1^2 + 2^2 ..... + (n-1)^2]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[ n+\frac{4}{n^2}\{(n-1)n(\frac{2n-1}{6}\}]$

$\displaystyle\sf =2 \lim_{n \to \infty}( \frac{1}{n} )[n + \frac{2}{3} \{\frac{(n-1)(2n-1)}{n} \}]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[n+\frac{2}{3}(1-\frac{1}{n} )(2-\frac{1}{n})]$

• $\sf n \to \infty , \frac{1}{n} \to 0$

$\displaystyle\sf \int_{0}^{2} ( {x}^{2} + 1)dx = 2 (1+ \frac{4}{3})$

$\displaystyle\boxed{\sf\red{ \int_{0}^{2} ( {x}^{2} + 1)dx = \frac{14}{3}}}$

Answer 2

$\bold \color{red} \star \underbrace \green{hence..the \: right \: answer \: is \: \frac{14}{3} } \star$

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