Find the area enclosed between the curves
$y = x^{2} - 5 x, y = 4 - 2 x$

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Detailed Solution

$y = x^{2} - 5 x, y = 4 - 2 x \Rightarrow x^{2} - 5 x = 4 - 2 x$

$\Rightarrow x^{2} - 3 x - 4 = 0 \Rightarrow (x - 4) (x + 1) = 0 \Rightarrow x = - 1, 4$

${\int_{- 1}^{4} (4 - 2 x - x^{2} + 5 x) d x = 4 x + \frac{3 x^{2}}{2} - \frac{x^{3}}{3}]}_{- 1}^{4}$

$= 16 + 24 - \frac{64}{3} + 4 - \frac{3}{2} - \frac{1}{3}$

$= 44 - \frac{65}{3} - \frac{3}{2} = \frac{264 - 130 - 9}{6} = \frac{125}{6} sq.unit$

$∴$ The area enclosed between the curves

$y = x^{2} - 5 x and y = 4 - 2 x is \frac{125}{6} sq.units$

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