Find the area of the closed figure bounded by the following curve
$y = x^{2} - 2 x + 3, y = 3 x - 1$ .

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Solution

Given the curves $y = x^{2} - 2 x + 3, y = 3 x - 1$
$3 x - 1 = x^{2} - 2 x + 3$
$x^{2} - 5 x + 4 = 0$
$x = 1, 4$
The area enclosed between $x = 1$ and $x = 4$ is $\int_{1}^{4} (3 x - 1 - (x^{2} - 2 x + 3)) d x$
$[- \frac{x^{3}}{3} + 5 \frac{x^{2}}{2} - 4 x]_{1}^{4} = - \frac{63}{3} + 5 (\frac{15}{2}) - 12 = \frac{9}{2} s q . u n i t s$

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