The area enclosed between the curves y^2 = 4x | vedclass

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(A) The given curves are $y^2 = 4x$ (a parabola opening to the right) and $y = |x|$ (a $V$-shaped graph). To find the points of intersection,we set $y

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$\frac{8}{3}$ sq. units

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(A) The given curves are $y^2 = 4x$ (a parabola opening to the right) and $y = |x|$ (a $V$-shaped graph). To find the points of intersection,we set $y^2 = x^2$ (since $y = |x| \implies y^2 = x^2$). Substituting $y^2 = 4x$ into $x^2 = y^2$,we get $x^2 = 4x$,which implies $x^2 - 4x = 0$,so $x(x - 4) = 0$. The points of intersection are $x = 0$ and $x = 4$. For $x \in [0, 4]$,the parabola $y = 2\sqrt{x}$ lies above the line $y = x$. The area $A$ is given by the integral: $A = \int_{0}^{4} (2\sqrt{x} - x) \, dx$ $A = [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2}]_{0}^{4}$ $A = [\frac{4}{3}x^{3/2} - \frac{x^2}{2}]_{0}^{4}$ $A = (\frac{4}{3} \cdot 8 - \frac{16}{2}) - (0 - 0)$ $A = \frac{32}{3} - 8 = \frac{32 - 24}{3} = \frac{8}{3}$ sq. units.

The area enclosed between the curves $y^2 = 4x$ and $y = |x|$ is

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