Math Problem Statement

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

Solution

To find the area of the region bounded by the curves y=x+1y = \sqrt{x+1}, y+x=1y + x = 1, and the xx-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=x+1y = \sqrt{x+1} and y+x=1y + x = 1 intersect.

Set y=x+1y = \sqrt{x+1} equal to y=1xy = 1 - x: x+1=1x\sqrt{x+1} = 1 - x

Square both sides to remove the square root: (x+1)2=(1x)2(\sqrt{x+1})^2 = (1 - x)^2 x+1=12x+x2x + 1 = 1 - 2x + x^2

Rearrange the equation to standard quadratic form: x23x=0x^2 - 3x = 0 x(x3)=0x(x - 3) = 0

So, x=0x = 0 or x=3x = 3.

Now, substitute these xx-values back into y=1xy = 1 - x to find the corresponding yy-values:

  • For x=0x = 0: y=10=1y = 1 - 0 = 1
  • For x=3x = 3: y=13=2y = 1 - 3 = -2

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

2. Setting up the integrals

We will split the integral into two parts:

  • From x=0x = 0 to x=1x = 1 (where x+1\sqrt{x+1} is above the xx-axis)
  • From x=1x = 1 to x=3x = 3 (where y+x=1y + x = 1 is above the xx-axis)

The area AA can be calculated as: A=01x+1dx+13(1x)dxA = \int_{0}^{1} \sqrt{x+1} \, dx + \int_{1}^{3} (1 - x) \, dx

3. Computing the integrals

First Integral:

01x+1dx\int_{0}^{1} \sqrt{x+1} \, dx

Let u=x+1u = x + 1. Then, du=dxdu = dx. When x=0x = 0, u=1u = 1. When x=1x = 1, u=2u = 2. 12udu=12u1/2du\int_{1}^{2} \sqrt{u} \, du = \int_{1}^{2} u^{1/2} \, du =[23u3/2]12= \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} =23(23/213/2)= \frac{2}{3} (2^{3/2} - 1^{3/2}) =23(221)= \frac{2}{3} (2\sqrt{2} - 1)

Second Integral:

13(1x)dx\int_{1}^{3} (1 - x) \, dx =[xx22]13= \left[ x - \frac{x^2}{2} \right]_{1}^{3} =[(392)(112)]= \left[ (3 - \frac{9}{2}) - (1 - \frac{1}{2}) \right] =(34.5)(10.5)= \left( 3 - 4.5 \right) - \left( 1 - 0.5 \right) =1.50.5= -1.5 - 0.5 =2= -2

4. Total Area

The total area AA is the sum of the areas of the two integrals: A=23(221)+(2)A = \frac{2}{3} (2\sqrt{2} - 1) + (-2) =42232= \frac{4\sqrt{2} - 2}{3} - 2 =42263= \frac{4\sqrt{2} - 2 - 6}{3} =4283= \frac{4\sqrt{2} - 8}{3}

So, the area of the region is: 4283\boxed{\frac{4\sqrt{2} - 8}{3}}

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

Here are 5 questions you might want to ask next:

  1. How do you find the volume of a solid of revolution?
  2. What is the method of partial fractions in calculus?
  3. Can you explain the Fundamental Theorem of Calculus?
  4. How do you solve a system of linear equations using matrices?
  5. What are the properties of definite integrals?

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

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Math Problem Analysis

Mathematical Concepts

Integration
Quadratic Equations
Area Calculation

Formulas

Integral calculus
Quadratic formula

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College