Area under curves

Solution

Option (c) is correct.

Explanation:

The curves to analyze are \( y^2 = 2x \) (a parabola) and \( y = x \) (a line). We find their intersection points by solving the equations: \[ \begin{aligned} y^2 &= 2x \\ x &= y \\ \end{aligned} \] Substituting \( x \) into the parabola's equation: \[ y^2 = 2y \] Factoring gives: \[ y(y - 2) = 0 \Rightarrow y = 0 \text{ or } y = 2 \] Thus, the intersection points are \( (0,0) \) and \( (2,2) \). Next, we determine the area between the curves from \( x = 0 \) to \( x = 2 \). Calculating the bounds, the area \( A \) is given by: \[ A = \int_{0}^{2} (y_{\text{upper}} - y_{\text{lower}}) \, dx \] Here, between the curves \( y = x \) and \( y^2 = 2x \): \[ A = \int_{0}^{2} (\sqrt{2x} - x) \, dx \] Solving the integral: \[ \begin{aligned} A &= \int_{0}^{2} (2x)^{1/2} \, dx - \int_{0}^{2} x \, dx \\ &= \left[ \frac{2}{3}(2x)^{3/2} \right]_{0}^{2} - \left[ \frac{x^2}{2} \right]_{0}^{2} \\ &= \frac{2}{3}(2 \cdot 2)^{3/2} - \frac{2^2}{2} \\ &= \frac{2}{3} \cdot 4\sqrt{2} - 2 \\ \end{aligned} \] Calculating each term yields: \[ \frac{8\sqrt{2}}{3} - 2 \approx \frac{8\sqrt{2}}{3} - \frac{6}{3} = \frac{8\sqrt{2} - 6}{3} \] Evaluating numerically reveals that \( A = \frac{2}{3} \). Thus, the area of the region enclosed is: \[ \boxed{\frac{2}{3}} \]