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Definite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\limits^3_{0}\big|3\text{x}-1\big|\text{dx}$
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Answer

$\int^\limits3_{0}\big|3\text{x}-1\big|\text{dx}=\int^\limits{\frac{1}{3}}_0-(3\text{x}-1)\text{dx}+\int^\limits{3}_\frac{1}{3}(3\text{x}-1)\text{dx}$
$=-\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^{\frac{1}{3}}_0+\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^3_{\frac{1}{3}}$
$=-\bigg[\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)-(0)\Big]+\bigg[\Big(\frac{3\times9}{2}-3\Big)-\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]+\bigg[\Big(\frac{27}{2}-3\Big)-\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(-\frac{1}{6}\Big)\bigg]+\bigg[10\frac{1}{2}+\frac{1}{6}\bigg]$
$=\frac{1}{6}+10\frac{1}{2}+\frac{1}{6}$
$=\frac{1}{3}+\frac{21}{2}$
$=\frac{2+63}{6}$
$=\frac{65}{6}$

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