Let `f(x)=min(x+1,sqrt(1-x))AA x le 1`. Then, the area (in sq. units( bounded by `y=f(x), y=0 and x=0` from `y=0` to `x=1` is equal to

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(4)/(9)`
D. 1

Answer

To find the area bounded by the curve \( y = f(x) \), the x-axis, and the lines \( x = 0 \) and \( x = 1 \), we first need to analyze the function \( f(x) \).

### Step 1: Define the function
The function is given as:
\[ f(x) = \min(x + 1, \sqrt{1 - x}) \quad \text{for } x \leq 1 \]

### Step 2: Find the intersection points
To determine where the two functions intersect, we set:
\[ x + 1 = \sqrt{1 - x} \]

### Step 3: Square both sides
Squaring both sides gives:
\[ (x + 1)^2 = 1 - x \]
Expanding this, we have:
\[ x^2 + 2x + 1 = 1 - x \]
Rearranging gives:
\[ x^2 + 3x = 0 \]

### Step 4: Factor the equation
Factoring out \( x \):
\[ x(x + 3) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = -3 \]

### Step 5: Determine the relevant intersection point
Since we are only interested in the interval \( x \leq 1 \), we will consider \( x = 0 \) as the relevant intersection point.

### Step 6: Evaluate \( f(x) \) in the interval [0, 1]
Next, we evaluate \( f(x) \) in the interval \( [0, 1] \):
- For \( x = 0 \):
\[ f(0) = \min(0 + 1, \sqrt{1 - 0}) = \min(1, 1) = 1 \]
- For \( x = 1 \):
\[ f(1) = \min(1 + 1, \sqrt{1 - 1}) = \min(2, 0) = 0 \]

### Step 7: Analyze the behavior of \( f(x) \)
- For \( 0 \leq x < 1 \), \( x + 1 \) is always greater than \( \sqrt{1 - x} \) until \( x \) approaches 1.
- Therefore, \( f(x) = \sqrt{1 - x} \) for \( 0 \leq x < 1 \).

### Step 8: Set up the integral for the area
The area \( A \) bounded by \( y = f(x) \), the x-axis, and the lines \( x = 0 \) and \( x = 1 \) can be expressed as:
\[ A = \int_0^1 f(x) \, dx = \int_0^1 \sqrt{1 - x} \, dx \]

### Step 9: Evaluate the integral
To evaluate the integral:
\[ \int \sqrt{1 - x} \, dx \]
We can use the substitution \( u = 1 - x \), thus \( du = -dx \) and changing limits:
- When \( x = 0 \), \( u = 1 \)
- When \( x = 1 \), \( u = 0 \)

The integral becomes:
\[ -\int_1^0 u^{1/2} \, du = \int_0^1 u^{1/2} \, du \]
Calculating this integral:
\[ = \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{2}{3} (1 - 0) = \frac{2}{3} \]

### Final Result
Thus, the area bounded by the curve, the x-axis, and the lines \( x = 0 \) and \( x = 1 \) is:
\[ \boxed{\frac{2}{3}} \text{ square units} \]