Question
Evaluate the following integrals:$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
✓
Answer
We know that,$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
$=\int^\limits\frac{1}{2}_{-1}-(2\text{x}+1)\text{dx}+\int\limits_{-\frac{1}{2}}^{1}(2\text{x}+1)\text{dx}$
$=-\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{-\frac{1}{2}}_{-1}+\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{1}_{-\frac{1}{2}}$
$=-\bigg[\Big(\frac{2}{8}-\frac{1}{2}\Big)-\Big(\frac{2}{2}-1\Big)\bigg]+\bigg[\Big(\frac{2}{2}+1\Big)-\Big(\frac{2}{8}-\frac{1}{2}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{4}-\frac{1}{2}\Big)-(1-1)\bigg]+\bigg[(1+1)+\Big(\frac{1}{4}-\frac{1}{2}\Big)\bigg]$
$=-\Big[-\frac{1}{4}\Big]++\Big[2+\frac{1}{4}\Big]$
$=\frac{1}{4}+2+\frac{1}{4}$
$=2\frac{1}{2}$
$=\int^\limits\frac{1}{2}_{-1}-(2\text{x}+1)\text{dx}+\int\limits_{-\frac{1}{2}}^{1}(2\text{x}+1)\text{dx}$
$=-\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{-\frac{1}{2}}_{-1}+\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{1}_{-\frac{1}{2}}$
$=-\bigg[\Big(\frac{2}{8}-\frac{1}{2}\Big)-\Big(\frac{2}{2}-1\Big)\bigg]+\bigg[\Big(\frac{2}{2}+1\Big)-\Big(\frac{2}{8}-\frac{1}{2}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{4}-\frac{1}{2}\Big)-(1-1)\bigg]+\bigg[(1+1)+\Big(\frac{1}{4}-\frac{1}{2}\Big)\bigg]$
$=-\Big[-\frac{1}{4}\Big]++\Big[2+\frac{1}{4}\Big]$
$=\frac{1}{4}+2+\frac{1}{4}$
$=2\frac{1}{2}$
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