The area bounded by the coordinate axes and the curve `sqrt(x) + sqrt(y) = 1`, is

Options

A. `1` square unit
B. `1/2` square unit
C. `1/3` square unit
D. `1/6` square unit

Explanation

To find the area bounded by the coordinate axes and the curve \(\sqrt{x} + \sqrt{y} = 1\), we will follow these steps:

### Step 1: Identify the intercepts of the curve
To find the points where the curve intersects the axes, we set \(x = 0\) and \(y = 0\).

1. **When \(x = 0\)**:
\[ \sqrt{0} + \sqrt{y} = 1 \implies \sqrt{y} = 1 \implies y = 1 \]
So, the curve intersects the y-axis at the point \((0, 1)\).

2. **When \(y = 0\)**:
\[ \sqrt{x} + \sqrt{0} = 1 \implies \sqrt{x} = 1 \implies x = 1 \]
So, the curve intersects the x-axis at the point \((1, 0)\).

### Step 2: Set up the integral
The area we want to calculate is bounded by the curve and the axes, which forms a right triangle with vertices at \((0, 0)\), \((0, 1)\), and \((1, 0)\).

To set up the integral, we will express \(y\) in terms of \(x\):
\[ \sqrt{y} = 1 - \sqrt{x} \implies y = (1 - \sqrt{x})^2 \]

### Step 3: Determine the limits of integration
The limits for \(x\) will be from \(0\) to \(1\).

### Step 4: Set up the area integral
The area \(A\) can be found using the integral:
\[ A = \int_{0}^{1} (1 - \sqrt{x})^2 \, dx \]

### Step 5: Expand the integrand
First, expand \((1 - \sqrt{x})^2\):
\[ (1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x \]
Thus, the integral becomes:
\[ A = \int_{0}^{1} (1 - 2\sqrt{x} + x) \, dx \]

### Step 6: Integrate term by term
Now we integrate each term:
\[ A = \int_{0}^{1} 1 \, dx - 2\int_{0}^{1} \sqrt{x} \, dx + \int_{0}^{1} x \, dx \]

Calculating each integral:
1. \(\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1\)
2. \(\int_{0}^{1} \sqrt{x} \, dx = \left[\frac{2}{3}x^{3/2}\right]_{0}^{1} = \frac{2}{3}\)
3. \(\int_{0}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2}\)

### Step 7: Substitute back into the area formula
Substituting these results back into the area formula:
\[ A = 1 - 2\left(\frac{2}{3}\right) + \frac{1}{2} \]
\[ A = 1 - \frac{4}{3} + \frac{1}{2} \]

### Step 8: Simplify the expression
To combine these fractions, we need a common denominator, which is \(6\):
\[ A = \frac{6}{6} - \frac{8}{6} + \frac{3}{6} = \frac{6 - 8 + 3}{6} = \frac{1}{6} \]

### Final Answer
Thus, the area bounded by the coordinate axes and the curve \(\sqrt{x} + \sqrt{y} = 1\) is:
\[ \boxed{\frac{1}{6}} \]