Question

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

Options

• `4/3`

• 1

• `2/3`

• `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
\[y^2 = x\text{ and }2y = x\]
\[ \Rightarrow y^2 = 2y \]
\[ \Rightarrow y^2 - 2y = 0\]
\[ \Rightarrow y = 0\text{ or }y = 2\]
\[ \Rightarrow x = 0\text{ or }x = 4\]
\[\text{ Thus O  }\left( 0, 0 \right)\text{ and A }\left( 4, 2 \right) \text{ are the points of intersection of the curve and straight line } . \]
Area bound by them]
\[A = \int_0^4 \left( y_1 - y_2 \right) dx .............\left[\text{Where, }y_1 = \sqrt{x}\text{ and }y_2 = \frac{x}{2} \right]\]
\[ = \int_0^4 \left( \sqrt{x} - \frac{x}{2} \right) dx\]
\[ = \left[ \frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{1}{2} \times \frac{x^2}{2} \right]_0^4 \]
\[ = \left[ \frac{2}{3} x^\frac{3}{2} - \frac{x^2}{4} \right]_0^4 \]
\[ = \frac{2}{3} 4^\frac{3}{2} - \frac{1}{4} \times 4^2 - 0\]
\[ = \frac{2}{3} \times 2^3 - \frac{16}{4}\]
\[ = \frac{16}{3} - 4\]
\[ = \frac{16 - 12}{3}\]
\[ = \frac{4}{3}\text{ sq units }\]