The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and x-axis is

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$\frac{2}{3}$ Point of intersection of the parabola x2 = 4y and straight line x = 2 is given by x^2 = 4y\text{ and }x = 2 \Rightarrow 4 = 4y \Rightarrow y = 1 A\left( 2, 1 \right)\text{ is the point of intersection of the curve and straight line } \text{ Area of shaded region OAB }= \int_0^2 y dx = \int_0^2 x^$\frac{2}{4}$ dx = \left[ x^$\frac{3}{12}$ \right]_0^2 = 2^$\frac{3}{12}$ - 0 = $\frac{2}{3}$\text{ square units } [Solution diagram: 44119_solution_1.png]

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