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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^2_{-1}\big(|\text{x}+1|+|\text{x}|+|\text{x}-1|\big)\text{dx}$ 
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Answer

We know that,
$\Rightarrow|\text{x}+1|=\begin{cases}\text{x}+1,&\text{ if }\text{ x}+1\geq0\\-(\text{x}+1),&\text{ if }\text{ x}+1<0\end{cases}\\=\begin{cases}\text{x}+1,&\text{ if }\text{ x}\geq-1\\-(\text{x}+1),&\text{ if }\text{ x}<-1\end{cases}$
$\Rightarrow|\text{x}|=\begin{cases}\text{x},&\text{ if }\text{ x}\geq0\\-\text{x},&\text{ if }\text{ x}<0\end{cases}$
$\Rightarrow|\text{x}-1|=\begin{cases}\text{x}-1,&\text{ if }\text{ x}-1\geq0\\-(\text{x}+1),&\text{ if }\text{ x}-1<0\end{cases}\\=\begin{cases}\text{x}-1,&\text{ if }\text{ x}\geq1\\-(\text{x}-1),&\text{ if }\text{ x}<1\end{cases}$
When $-1\leq\text{x}\leq0,$
$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+(-\text{x})+\big[-(\text{x}-1)\big]\\=2-\text{x}$
When $0\leq\text{x}\leq1,$
$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+\text{x}+\big[-(\text{x}-1)\big]\\=\text{x}+2$
When $1\leq\text{x}\leq2,$
$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+\text{x}+\text{x}-1\\=3\text{x}$
$\therefore\ \int\limits^2_{-1}\big(|\text{x}+1|+|\text{x}|+|\text{x}-1|\big)\text{dx}$
$=\int\limits^0_{-1}(2-\text{x})\text{dx}+\int\limits^1_{0}(\text{x}+2)\text{dx}+\int\limits^2_{1}3\text{x dx}$
$=\bigg[\frac{(2-\text{x})^2}{2\times(-1)}\bigg]^0_{-1}+\bigg[\frac{(\text{x}+2)^2}{2}\bigg]^1_0+3\times\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=-\frac{1}{2}(4-9)+\frac{1}{2}(9-4)+\frac{3}{2}(4-1)$
$=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}$
$=\frac{19}{2}$

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