Sketch the region bounded by the given lines. Then express the​ region's area as an iterated double integral and evaluate the integral.
The coordinate axes and the line nothingxplus2yequals2.


### Step 1: Analyze the line equation
The given equation is \( x + 2y = 2 \). To understand the geometry of the region bounded by this line and the coordinate axes, we first express the equation in slope-intercept form.

Solve for \( y \):
\[
x + 2y = 2 \implies 2y = 2 - x \implies y = 1 - \frac{x}{2}
\]

This is the equation of a straight line with slope \( -\frac{1}{2} \) and y-intercept at \( y = 1 \).

### Step 2: Find the points of intersection
To determine the region bounded by the line and the coordinate axes, we find the intercepts.

- **x-intercept**: Set \( y = 0 \) in the equation \( x + 2y = 2 \):
  \[
  x + 2(0) = 2 \implies x = 2
  \]
  So, the line intersects the x-axis at \( (2, 0) \).

- **y-intercept**: Set \( x = 0 \) in the equation \( x + 2y = 2 \):
  \[
  0 + 2y = 2 \implies y = 1
  \]
  So, the line intersects the y-axis at \( (0, 1) \).

### Step 3: Sketch the region
The region is bounded by the line \( x + 2y = 2 \), the x-axis, and the y-axis. It forms a triangular region with vertices at:
1. \( (0, 0) \) (the origin)
2. \( (2, 0) \) (x-intercept)
3. \( (0, 1) \) (y-intercept)

### Step 4: Set up the double integral
The region is described by:
- \( x \) ranges from 0 to 2.
- For each fixed \( x \), \( y \) ranges from 0 to the line \( y = 1 - \frac{x}{2} \).

The area of the region can be expressed as the following double integral:
\[
A = \int_0^2 \int_0^{1 - \frac{x}{2}} dy \, dx
\]

### Step 5: Evaluate the double integral
First, evaluate the inner integral with respect to \( y \):
\[
\int_0^{1 - \frac{x}{2}} dy = \left[ y \right]_0^{1 - \frac{x}{2}} = 1 - \frac{x}{2}
\]

Now, substitute this into the outer integral and evaluate with respect to \( x \):
\[
A = \int_0^2 \left( 1 - \frac{x}{2} \right) dx
\]

Split the integral:
\[
A = \int_0^2 1 \, dx - \int_0^2 \frac{x}{2} \, dx
\]

The first integral:
\[
\int_0^2 1 \, dx = x \Big|_0^2 = 2
\]

The second integral:
\[
\int_0^2 \frac{x}{2} \, dx = \frac{1}{2} \int_0^2 x \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^2 = \frac{1}{2} \times \frac{4}{2} = 1
\]

Now subtract:
\[
A = 2 - 1 = 1
\]

### Final Answer:
The area of the region is \( 1 \) square unit.

---

Would you like further clarification on any steps, or help visualizing the region? 

Here are some related questions you might want to explore:
1. How do you find the equation of a line given two points?
2. What is the geometric meaning of a double integral in the context of areas?
3. How do you set up bounds for double integrals in different regions?
4. What are other methods for calculating areas of regions in the plane?
5. How would the area change if the line was shifted or rotated?

**Tip**: Double integrals can be applied to find volumes as well, by adding functions inside the integrand instead of just 1!

Sketch the region bounded by the given lines. Then express the region's area as an iterated double integral and evaluate the integral. The coordinate axes and the line x + 2y = 2.

Learn how to sketch the region bounded by the line x + 2y = 2 and the coordinate axes, and express the area of the region as an iterated double integral. This problem covers analytic geometry, setting up double integrals, and evaluating the area. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation