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INTEGRALS — MATHS STD 12 Science — Question

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

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