Integrals — Class 12 Maths Solution

Let $f\left( x \right) = {x^2}$ $\Rightarrow$ $\int\limits_2^3 {f\left( x \right)\,dx} = \int\limits_2^3 {{x^2}} dx$

Where, $a = 2,b = 3,nh = 1$

Now, $f\left( a \right) = f\left( 2 \right) = {2^2}$

$f\left( {a + h} \right) = f\left( {2 + h} \right) = {(2 + h)^2} = 4 + {h^2} + 4h$

$f\left( {a + 2h} \right) = f\left( {2 + 2h} \right) = {(2 + 2h)^2} = 4 + 4{h^2} + 8h$

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

$= 4 + {\left( {n - 1} \right)^2}{h^2} + 4\left( {n - 1} \right)h$

By definition $\int\limits_a^b {f\left( x \right)} dx$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( a \right) + f\left( {a + h} \right) + f\left( {a + 2h} \right) + ... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$

$\therefore$ $\int\limits_2^3 {{x^2}} dx$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {4n + {h^2}\left( {1 + 4 + 9 + ...... + {{\left( {n - 1} \right)}^2}} \right) + 4h\left[ {1 + 2 + ... + \left( {n - 1} \right)} \right]} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {4n + \cfrac{{{h^2}\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6} + 4h\cfrac{{\left( {n - 1} \right)n}}{2}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {4nh + \cfrac{{\left( {nh - h} \right)\left( {nh} \right)\left( {2nh - h} \right)}}{6} + 2\left( {nh - h} \right)nh} \right]$

$= 4 \times 1 + \cfrac{{1 \times 2}}{6} + 2 \times 1 \times 1 = 4 + \cfrac{1}{3} + 2 = 6 + \cfrac{1}{3} = \cfrac{{19}}{3}$