Area of the region bounded by the curve \(y = \sqrt{x}\) and lines \(x + y = 2, y = 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{7}{6}\) square units
The curve meets the line where \(\sqrt{x} = 2 - x\):
\(x = 4 - 4x + x^2\)
\(x^2 - 5x + 4 = 0,\quad x = 1 \text{ or } 4\)
x = 4 gives \(2 - x = -2 \lt 0\), so it is rejected. The curve and the line meet at (1, 1). The line meets the x-axis at (2, 0), and the curve starts at (0, 0).
So the region has corners (0, 0), (1, 1) and (2, 0): it lies under the curve from x = 0 to x = 1 and under the line from x = 1 to x = 2. Splitting the area at x = 1:
\( \displaystyle A = \int_{0}^{1}\sqrt{x}\,dx + \int_{1}^{2}(2 - x)\,dx \)
\( \displaystyle = \left[\frac{2}{3}x^{3/2}\right]_{0}^{1} + \left[2x - \frac{x^2}{2}\right]_{1}^{2} \)
\( \displaystyle = \frac{2}{3} + \left(2 - \frac{3}{2}\right) = \frac{2}{3} + \frac{1}{2} = \frac{7}{6} \)
Shortcut: Integrate along y. For \(0 \le y \le 1\), x runs from the curve \(x = y^2\) to the line \(x = 2 - y\):
\( \displaystyle A = \int_{0}^{1}(2 - y - y^2)\,dy = 2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6} \)