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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Evaluate $\int_{2}^{3} x^{2} d x$
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Answer

f(x) is continuous in [2, 3]
$\int_{a}^{b} f(x) d x$ = $\mathop {\lim }\limits_{n \to \infty } h\sum\limits_{r = 0}^{n - 1} f (a + rh)$ where $\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$ 
Here,
$\int_{2}^{3}\left(x^{2}\right) d x$ = $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} f \left( {2 + \left( {\frac{r}{n}} \right)} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {{{\left( {2 + \left( {\frac{r}{n}} \right)} \right)}^2}}$ 
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\frac{{{r^2}}}{{{n^2}}} + 4 + \frac{{4r}}{n}} \right)}$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{(n - 1)(n)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{4(n - 1)(n)}}{{2n}}} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {{n^2} - n} \right)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{2\left( {{n^2} - n} \right)}}{n}} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left(\frac{\left(2 n^{3}-2 n^{2}-n^{2}+n\right)}{6 n^{2}}+4 n+\frac{2\left(n^{2}-n\right)}{n}\right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {2{n^3} - 3{n^2} + n} \right) + \left( {24{n^3}} \right) + \left( {12{n^3} - 12{n^2}} \right)}}{{6{n^2}}}} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^2}}}} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^3}}}} \right)$ 
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38}}{6}} \right) - \left( {\frac{{15}}{{6n}}} \right) + \left( {\frac{1}{{6{n^2}}}} \right)$ 
= $\frac{38}{6}$ 
= $\frac{19}{3}$ 

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