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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the definite integral in Exercise:
$\int^{4}_{1}[\text{x}-1|+|\text{x}-2|+|\text{x}-3|]\text{dx}$
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Answer

$\text{Let I}=\int^{4}_{1}[\text{x}-1|+|\text{x}-2|+|\text{x}-3|]\text{dx}$
$\Rightarrow\text{I}=\int^{4}_{1}\text{x}-1|\text{dx}+\int^{4}\limits_{1}|\text{x}-2|\text{dx}+\int^{4}\limits_{1}|\text{x}-3|\text{dx}$
$\text{I}=\text{I}_{1}+\text{I}_{2}+\text{I}_{3}$
where,$\text{I}_{1}=\int^{4}\limits_{1}\text{x}-1|\text{dx},\text{I}_{1}=\int^{4}\limits_{1}|\text{x}-2|\text{dx},$and $\text{I}_{3}=\int^{4}\limits_{1}\text{x}-3|\text{dx}$
$\text{I}_{1}=\int^{4}_{1}|\text{x}-1|\text{dx}$
$(\text{x}-1)\geq0$ for $1\le\text{x}\le4$
$\therefore\text{I}_{1}=\int^{4}\limits_{1}(\text{x}-1)\text{dx}$
$\Rightarrow\text{I}_{1}=\bigg[\frac{\text{x}^{2}}{\text{2}}-\text{x}\bigg]^{4}_{1}$
$\Rightarrow\text{I}_{1}=\bigg[8-4-\frac{1}{2}+1\bigg]=8-4+\frac{1}{2}$
$\text{I}_{2}=\int^{4}\limits_{1}|\text{x}-2|\text{dx}$
$\text{x}-2\ge0$ for $2\le\text{x}\le4$ and $\text{x}-2\le0$ for $1\le\text{x}\le2$
$\therefore\ \text{I}_{2}=\int^{2}\limits_{1}(2-\text{x)}\text{dx}+\int^{4}\limits_{2}(\text{x}-2)\text{dx}$
$\Rightarrow\ \text{I}_{2}=\bigg[2\text{x}-\frac{\text{x}^{2}}{2}\bigg]^{2}_{1}+\bigg[\frac{\text{x}^{2}}{2}-2\text{x}\bigg]^{4}_{2}$
$\Rightarrow\ \text{I}_{2}=\bigg[4-2-2+\frac{1}{2}\bigg]+\bigg[8-8-2+4\bigg]$
$\Rightarrow\ \text{I}_{2}=\frac{1}{2}+2=\frac{5}{2}$
$\text{I}_{3}=\int^{4}\limits_{1}|\text{x}-3|\text{dx}$
$\text{x}-3\ge0$ for $3\le\text{x}\le4$ and $\text{x}-3\le$ for $1\le\text{x}\le3$
$\therefore\text{I}_{3}=\int^{3}\limits_{1}(3-\text{x})\text{dx}+\int^{4}\limits_{3}(\text{x}-3)\text{dx}$
$\Rightarrow\ \text{I}_{3}=\bigg[3\text{x}-\frac{\text{x}^{2}}{2}\bigg]^{3}_{1}+\bigg[\frac{\text{x}^{2}}{2}-3\text{x}\bigg]^{4}_{3}$
$\Rightarrow\ \text{I}_{3}=\bigg[9-\frac{9}{2}-3+\frac{1}{2}\bigg]+\bigg[8-12-\frac{9}{2}+9\bigg]$
$\Rightarrow\ \text{I}_{3}=\big[6-4\big]+\bigg[\frac{1}{2}\bigg]=\frac{5}{2}$
From equation (1),(2),(3), and (4), we obtain
$\text{I}=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}$

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