The area between the parabola y = x^2 and the | vedclass

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(A) Given curves are $y = x^2$ and $y = x$.To find the points of intersection,set $x^2 = x$,which gives $x^2 - x = 0$,so $x(x - 1) = 0$.Thus,the point

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$\frac{1}{6} 	ext{ sq. unit}$

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(A) Given curves are $y = x^2$ and $y = x$.To find the points of intersection,set $x^2 = x$,which gives $x^2 - x = 0$,so $x(x - 1) = 0$.Thus,the points of intersection are $x = 0$ and $x = 1$.In the interval $[0, 1]$,the line $y = x$ lies above the parabola $y = x^2$.Therefore,the required area $A$ is given by:$A = \int_0^1 (x - x^2) dx$$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} ight]_0^1$$A = \left( \frac{1}{2} - \frac{1}{3} ight) - (0 - 0)$$A = \frac{3 - 2}{6} = \frac{1}{6} 	ext{ sq. unit}$.

The area between the parabola $y = x^2$ and the line $y = x$ is

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