Integrals — Class 12 Maths Solution

Let $f\left( x \right) = x,$ then $\int\limits_a^b {\left( f \right)x\,dx} = \int\limits_a^b {x\,dx}$

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

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

Since $\int\limits_a^b {f\left( x \right)\,dx}$

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

Where ,$nh = b - a$
$\therefore$ $I = \int\limits_a^b x dx$

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

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

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

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

$= \cfrac{{2a\left( {b - a} \right) + \left( {b - a} \right)\left( {b - a} \right)}}{2} = \cfrac{{\left( {2a + b - a} \right)\left( {b - a} \right)}}{2}$

$= \cfrac{{\left( {b + a} \right)\left( {b - a} \right)}}{2} = \cfrac{{{b^2} - {a^2}}}{2}$