The area bounded by the parabola y^2=4 x and the line 2 x-3 y+4=0 ,...

Intersecting points are $x=1,4$ \begin{aligned} & \therefore \text { Required area }=\int_1^4\left[2 \sqrt{x}-\left(\frac{2 x+4}{3}\right)\right] d x \\ & =\left.\frac{2 x^{3 / 2}}{3 / 2}\right|_1 ^4-\left.\frac{2 x^2}{3 \times 2}\right|_1 ^4-\left.\frac{4}{3} x\right|_1 ^4 \end{aligned}\begin{aligned} & =\frac{4}{3}\left(4^{3 / 2}-1^{3 / 2}\right)-\frac{1}{3}(16-1)\left[\frac{4}{3}(4)-\frac{4}{3}\right] \\ & =\frac{4}{3}(7)-5-4=\frac{28}{3}-9=\frac{28-27}{3}=\frac{1}{3} \end{aligned}