Find the area of the closed figure bounded by the following curve
$y$ = $x^{2}$ - $2 x + 2, y = 2 + 4 x$ - $x^{2}$ .

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Solution

Given the curves $y = x^{2} - 2 x + 2 a n d y = 2 + 4 x - x^{2}$
$x^{2} - 2 x + 2 = 2 + 4 x - x^{2}$
$2 x^{2} - 6 x = 0$
$x (x - 3) = 0 \Rightarrow x = 0, 3$
The area enclosed between curve ,x=0 and x=3 is $\int_{0}^{3} (- x^{2} + 4 x + 2 - (x^{2} - 2 x + 2)) d x$
$\int_{0}^{3} (- 2 x^{2} + 6 x) d x = [- 2 \frac{x^{3}}{3} + 3 x^{2}]_{0}^{3} = 27 - 18 = 9 s q . u n i t s$

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