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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate $\int\limits_0^{3}(2x^2 + 3x + 5)dx$ as limit of a sum.
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Answer

Here h = $\frac{3}{\text{n}}$ and $f(x) = 2x^2 + 3x + 5$
$\therefore$ I = $\lim\limits_{\text{ h} \to 0}\text{h}\cdot[\text{f(0) + f(h) + f(2h) + f(3h) +.......+ f}\left\{\overline{\text{(n - 1)}}\text{h}\right\}]$
=$\lim\limits_{\text{h} \to 0}\\_{\text{h} \to \infty}$ $\frac{3}{\text{n}}\cdot$[$(5) + (2h^2+ 3h + 5) + (2.2^2h^2+ 3.2h + 5) + .........+ {2(n - 1)^2 h^2 + 3 (n - 1) h + 5 }]$
=$\lim\limits_{\text{n} \to \infty}$ $\frac{3}{\text{n}}\cdot$ [$(5 + 5 + 5 +.........n $terms) $+ 2h^2{1^2+ 2^2+ 3^2+ ........+ (n - 1)^2} + 3h {1 + 2 + 3 + ..... (n - 1)}]$
= $\lim\limits_{\text{n} \to \infty}$ $\frac{3}{\text{n}}\cdot$ $\Bigg[\text{5n+2}\cdot\frac{9}{\text{n}^{2}}\cdot\text{n}\frac{\text{(n - 1)(2n - 1)}}{6}+\frac{3.3}{\text{n}}\cdot\frac{\text{n(n - 1)}}{2}\Bigg]$
= $\lim\limits_{\text{n} \to \infty}$ $3\Bigg[5+3\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Bigg]$
= $3\Big[5+6+\frac{9}{2}\Big]=\frac{93}{2}.$

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