The area enclosed between the curves $y = x$ and $y = 2x - x^{2}$ is (in square unit)

Question

Dr. Abubacker · Accepted Answer

(b)
The intersection points of given curves are (0,0) and (1,1)
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$	herefore Required\ area = \int_{0}^{1}{\left\lbrack \left( 2x - x^{2} ight) - x ightbrack dx}$
$= \int_{0}^{1}{\left( x - x^{2} ight)dx = \left\lbrack \frac{x^{2}}{2} - \frac{x^{3}}{3} ightbrack_{0}^{1}}$
$= \frac{1}{6}sq\ unit\$

Q. The area enclosed between the curves $y = x$ and $y = 2x - x^{2}$ is (in square unit)

Explanation

Reviewed by Dr. Abubacker

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