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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{1}\big(3\text{x}^2+1\text{x}\big)\text{dx}$
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Answer

We have,$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx} = \lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})+(\text{a}+2\text{h})+\\...+\text{f}[(\text{a}+(\text{n}-1)\text{h})]\}$
Here, $\text{a}=1, \text{b}=3\text{ f}(\text{x})=3\text{x}^2+1$ and $\text{h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}\Rightarrow \text{nh}=2$$\therefore\int\limits^3_1(3\text{x}^2+1)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{\text{f}(1)+\text{f}(1+\text{h})+\text{f}(1+2\text{h})...+\text{f}[1+(\text{n}-1)\text{h}]\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{[3\times1^2+1]+[3\times(1+\text{h})^2+1]\\+[3\times(1+2\text{h})^2+1]+.....+[3\times(1+(\text{n}-1)\text{h}^2+1]$
$=\lim\limits\text{h}\{3[1+(1+2\text{h}+\text{h}^2)+(1+4\text{h}+2^2\text{h}^2)+\\...+(1+2(\text{n}-1)\text{h}+(\text{n}-1)^2\text{h}^2)]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{3[\text{n}+2(1+2+...+(\text{n}-1))\text{h}+(1^2+2^2+\\.....+(\text{n}-1)^2)\text{h}^2]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+6\times\frac{\text{n}(\text{n}-1)}{2}\text{h}+3\times\frac{(\text{n}-1)\text{n}(2\text{n}-1)}{6}\text{h}^2\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+6\times\frac{\text{nh}(\text{nh}-\text{h)}}{2}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h}}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+3\times\text{nh}(\text{nh}-\text{h}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h})}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\times2+3\times2\times(2-\text{h})+3\times\frac{(2-\text{h)}\times2\times(2\times2-\text{h})}{6}\bigg]$
$= 8 + 6\times(2-0)+\frac{(2-0)\times2\times(4-0)}{2}$
$= 8+12+8$
$= 28$

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