Question

The area of region in the first quadrant that is bounded by $y=\sqrt{x}$, $y=2-x$ and x-axis is

• A. $\frac{5}{6}$
• B. $\frac{2}{3}$
• C. 1
• D. $\frac{7}{6}$

Hint:

Whenever a region is bounded by two "upper" functions, switching to integration with respect to $y$ (horizontal strips) usually allows you to solve the problem with a single integral instead of two.

Answer 1

The Correct Option is D

Area of the Region Bounded by Curves

Step 1: Find intersection points
The curves are:
\(y = \sqrt{x}\) and \(y = 2 - x\).
Set them equal: \(\sqrt{x} = 2 - x\)
Square both sides: \(x = (2 - x)^2 \implies x = 4 - 4x + x^2 \implies x^2 - 5x + 4 = 0\)
Factor: \((x - 4)(x - 1) = 0\)
- \(x = 4\) gives \(y = 2\) (outside first quadrant for \(y=2-x=-2\)), discard.
- \(x = 1\) gives \(y = 1\).
Intersection point: \((1,1)\)

Step 2: Determine the region
The region lies in the first quadrant, bounded:
- Below by \(y = 0\) (x-axis)
- Left by \(y = \sqrt{x} \implies x = y^2\)
- Right by \(y = 2 - x \implies x = 2 - y\)
- \(y\) ranges from 0 to 1

Step 3: Express area as an integral with respect to \(y\)
\[ \text{Area} = \int_{0}^{1} [(2 - y) - y^2] \, dy \]
Evaluate the integral:
\[ \int_{0}^{1} (2 - y - y^2) \, dy = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 \]
\[ = 2 - \frac{1}{2} - \frac{1}{3} = \frac{12 - 3 - 2}{6} = \frac{7}{6} \]

Step 4: Final Answer
The area of the region is \(\frac{7}{6}\), which corresponds to Option (4).