Find the area of the region bounded by y=\sqrt(x+1),y+x=1 and the x -axis.

To find the area of the region bounded by the curves \( y = \sqrt{x+1} \), \( y + x = 1 \), and the \( x \)-axis, we need to determine the points of intersection and then set up the appropriate integrals.

### 1. Finding the points of intersection
First, we need to find the points where \( y = \sqrt{x+1} \) and \( y + x = 1 \) intersect.

Set \( y = \sqrt{x+1} \) equal to \( y = 1 - x \):
\[ \sqrt{x+1} = 1 - x \]

Square both sides to remove the square root:
\[ (\sqrt{x+1})^2 = (1 - x)^2 \]
\[ x + 1 = 1 - 2x + x^2 \]

Rearrange the equation to standard quadratic form:
\[ x^2 - 3x = 0 \]
\[ x(x - 3) = 0 \]

So, \( x = 0 \) or \( x = 3 \).

Now, substitute these \( x \)-values back into \( y = 1 - x \) to find the corresponding \( y \)-values:
- For \( x = 0 \): \( y = 1 - 0 = 1 \)
- For \( x = 3 \): \( y = 1 - 3 = -2 \)

Since \( y = \sqrt{x+1} \) only provides non-negative \( y \)-values, the relevant intersection points are at \( (0, 1) \) and \( (3, -2) \).

### 2. Setting up the integrals
We will split the integral into two parts:
- From \( x = 0 \) to \( x = 1 \) (where \( \sqrt{x+1} \) is above the \( x \)-axis)
- From \( x = 1 \) to \( x = 3 \) (where \( y + x = 1 \) is above the \( x \)-axis)

The area \( A \) can be calculated as:
\[ A = \int_{0}^{1} \sqrt{x+1} \, dx + \int_{1}^{3} (1 - x) \, dx \]

### 3. Computing the integrals
#### First Integral:
\[ \int_{0}^{1} \sqrt{x+1} \, dx \]

Let \( u = x + 1 \). Then, \( du = dx \). When \( x = 0 \), \( u = 1 \). When \( x = 1 \), \( u = 2 \).
\[ \int_{1}^{2} \sqrt{u} \, du = \int_{1}^{2} u^{1/2} \, du \]
\[ = \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} \]
\[ = \frac{2}{3} (2^{3/2} - 1^{3/2}) \]
\[ = \frac{2}{3} (2\sqrt{2} - 1) \]

#### Second Integral:
\[ \int_{1}^{3} (1 - x) \, dx \]
\[ = \left[ x - \frac{x^2}{2} \right]_{1}^{3} \]
\[ = \left[ (3 - \frac{9}{2}) - (1 - \frac{1}{2}) \right] \]
\[ = \left( 3 - 4.5 \right) - \left( 1 - 0.5 \right) \]
\[ = -1.5 - 0.5 \]
\[ = -2 \]

### 4. Total Area
The total area \( A \) is the sum of the areas of the two integrals:
\[ A = \frac{2}{3} (2\sqrt{2} - 1) + (-2) \]
\[ = \frac{4\sqrt{2} - 2}{3} - 2 \]
\[ = \frac{4\sqrt{2} - 2 - 6}{3} \]
\[ = \frac{4\sqrt{2} - 8}{3} \]

So, the area of the region is:
\[ \boxed{\frac{4\sqrt{2} - 8}{3}} \]

If you have any questions or need further details, let me know!

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**Tip:** When solving integrals, always consider whether you can simplify the problem using substitution or by breaking the integral into simpler parts.

Learn how to calculate the area of the region bounded by the curves y=sqrt(x+1), y+x=1, and the x-axis with a detailed step-by-step solution. Includes integration techniques and quadratic equations. Suitable for college-level calculus students.

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