The area bounded by the parabola y^2 = 4x and | vedclass

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(B) Given equations are $y^2 = 4x$ and $2x - 3y + 4 = 0$.From the line equation,$x = \frac{3y - 4}{2}$.Substituting this into the parabola equation: $

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$\frac{1}{3}$

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(B) Given equations are $y^2 = 4x$ and $2x - 3y + 4 = 0$.From the line equation,$x = \frac{3y - 4}{2}$.Substituting this into the parabola equation: $y^2 = 4 \left( \frac{3y - 4}{2} \right) = 2(3y - 4) = 6y - 8$.$y^2 - 6y + 8 = 0 \implies (y - 2)(y - 4) = 0$.So,$y = 2$ and $y = 4$.When $y = 2$,$x = \frac{3(2) - 4}{2} = 1$. When $y = 4$,$x = \frac{3(4) - 4}{2} = 4$.The area is given by $\int_{2}^{4} (x_{line} - x_{parabola}) dy = \int_{2}^{4} \left( \frac{3y - 4}{2} - \frac{y^2}{4} \right) dy$.$= \left[ \frac{3y^2}{4} - 2y - \frac{y^3}{12} \right]_{2}^{4}$.$= \left( \frac{3(16)}{4} - 2(4) - \frac{64}{12} \right) - \left( \frac{3(4)}{4} - 2(2) - \frac{8}{12} \right)$.$= \left( 12 - 8 - \frac{16}{3} \right) - \left( 3 - 4 - \frac{2}{3} \right) = \left( 4 - \frac{16}{3} \right) - \left( -1 - \frac{2}{3} \right) = -\frac{4}{3} + \frac{5}{3} = \frac{1}{3}$ square units.

The area bounded by the parabola $y^2 = 4x$ and the line $2x - 3y + 4 = 0$,in square units,is

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