Evaluate: _ -1 ^1|5 x-3| d x - Vidyadip

Question

Evaluate: $\int_{-1}^1|5 x-3| d x$

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Answer

$\text { Let } I =\int_{-1}^1|5 x-3| d x$
$|5 x-3|=-(5 x-3) \text { when }(5 x-3)<0 \text { i.e. } x<\frac{3}{5}$
$=5 x-3 \text { when }(5 x-3)>0 \text { i.e., } x>\frac{3}{5}$
$\therefore I =\int_{-1}^{\frac{3}{5}}|5 x-3| d x+\int_{\frac{3}{5}}^1|5 x-3| d x$
$=\int_{-1}^{\frac{3}{5}}-(5 x-3) d x+\int_{\frac{3}{5}}^1(5 x-) d x$
$=-5 \int_{-1}^{\frac{3}{5}} x d x+3 \int_{-1}^{\frac{3}{5}} d x+5 \int_{\frac{3}{5}}^1 x d x-3 \int_{\frac{3}{5}}^1 d x$
$=-\frac{5}{2}\left[\frac{x^2}{2}\right]_{-1}^{\frac{3}{5}}+3[x]_{-1}^{\frac{3}{5}}+5\left[\frac{x^2}{2}\right]_{\frac{3}{5}}^1-3[x]_{\frac{3}{5}}^1$
$=-\frac{5}{2}\left[\left(\frac{3}{5}\right)^2-(-1)^2\right]+3\left[\frac{3}{5}-(-1)\right]+\frac{5}{2}\left[(1)^2-\left(\frac{3}{2}\right)^2\right]-3\left(1-\frac{3}{5}\right)$
$=\frac{5}{2}\left(\frac{9}{25}-1\right)+3\left(\frac{3}{5}+1\right)+\frac{5}{2}\left(1-\frac{9}{25}\right)-3\left(\frac{2}{5}\right)$
$=-\frac{5}{2}\left(\frac{-16}{25}\right)+3\left(\frac{8}{5}\right)+\frac{5}{2}\left(\frac{16}{25}\right)-\frac{6}{5}$
$=\frac{8}{5}+\frac{24}{5}+\frac{8}{5}-\frac{6}{5}$
$=\frac{34}{5}$

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