Integrals — Class 12 Maths Solution

Let $I = \int_0^2 {\left( {{x^2} + 3} \right)} dx$

Here, $a = 0,b = 2$ and $h = \frac{{b - a}}{n} = \frac{{2 - 0}}{n}$

$\Rightarrow$ $h = \frac{2}{n} \Rightarrow nh = 2 \Rightarrow f(x) = \left( {{x^2} + 3} \right)$

Now, $\int_0^2 {\left( {{x^2} + 3} \right)} dx = \mathop {\lim }\limits_{h \to 0} h[f(0) + f(0 + h) + f(0 + 2h) + \ldots + f\{ 0 + (n - 1)h\} ]$….(i)

$\Rightarrow$ $f(0 + h) = {h^2} + 3,f(0 + 2h) = 4{h^2} + 3 = {2^2}{h^2} + 3$

$f[0 + (n - 1)h] = \left( {{n^2} - 2n + 1} \right)h + 3 = {(n - 1)^2}h + 3$

From Equation (i), $\int_0^2 {\left( {{x^2} + 3} \right)} dx = \mathop {\lim }\limits_{h \to 0} h\left[ {3 + {h^2} + 3 + {2^2}{h^2} + 3 + {3^2}{h^2} + 3 + \ldots + {{(n - 1)}^2}{h^2} + 3} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {3n + {h^2}\left\{ {{1^2} + {2^2} + \ldots + {{(n - 1)}^2}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {3n + {h^2}\left( {\frac{{\left( {{n^2} - n} \right)(2n - 1)}}{6}} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {3n + \frac{{{h^2}}}{6}\left( {2{n^3} - {n^2} - 2{n^2} + n} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {3nh + \frac{{2{n^3}{h^3} - 3{n^2}{h^2} \cdot h + nh \cdot {h^2}}}{6}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {3 \cdot 2 + \frac{{2 \cdot 8 - 3 \cdot {2^2} \cdot h + 2 \cdot {h^2}}}{6}} \right] = \mathop {\lim }\limits_{h \to 0} \left[ {6 + \frac{{16 - 12h + 2{h^2}}}{6}} \right]$

$= 6 + \frac{{16}}{6} = 6 + \frac{8}{3} = \frac{{26}}{3}$