Question

Area bounded by parabola y² = x and straight line 2y = x is _________ .

पर्याय

• ${\displaystyle \frac{4}{3}}$

• 1

• ${\displaystyle \frac{2}{3}}$

• ${\displaystyle \frac{1}{3}}$

Answer 1

Point of intersection is obtained by solving the equation of parabola y² = x and equation of line 2y= x, we have
$$y^2=x\text{ and }2y=x$$
$$\Rightarrow y^2=2y$$
$$\Rightarrow y^2-2y=0$$
$$\Rightarrow y=0\text{ or }y=2$$
$$\Rightarrow x=0\text{ or }x=4$$
$$\text{ Thus O }(0,0)\text{ and A }(4,2)\text{ are the points of intersection of the curve and straight line }.$$
Area bound by them]
$$A={\int }_0^4(y_1-y_2)dx.............[\text{Where, }{y}_1=\sqrt{x}\text{ and }{y}_2=\frac{x}{2}]$$
$$={\int }_0^4(\sqrt{x}-\frac{x}{2})dx$$
$$={[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\times \frac{x^2}{2}]}_0^4$$
$$={[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^2}{4}]}_0^4$$
$$=\frac{2}{3}{4}^{\frac{3}{2}}-\frac{1}{4}\times 4^2-0$$
$$=\frac{2}{3}\times 2^3-\frac{16}{4}$$
$$=\frac{16}{3}-4$$
$$=\frac{16-12}{3}$$
$$=\frac{4}{3}\text{ sq units }$$