Use integration to find the area of the region enclosed by the graphs of
y
=
√
x
+
2
,
x
=
−
2
, and
y
=
3
.

Round your result to 2 decimal place.

To find the area of the region enclosed by the curves \( y = \sqrt{x} + 2 \), the vertical line \( x = -2 \), and the horizontal line \( y = 3 \), we need to set up an integral that will calculate this area.

### Step 1: Determine the Points of Intersection
First, we find where the curve \( y = \sqrt{x} + 2 \) intersects the line \( y = 3 \).

Set \( y = 3 \) in the equation \( y = \sqrt{x} + 2 \):
\[
3 = \sqrt{x} + 2
\]
Solve for \( x \):
\[
\sqrt{x} = 1 \quad \Rightarrow \quad x = 1
\]
So, the curves intersect at \( (1, 3) \).

### Step 2: Set Up the Integral
We will integrate with respect to \( x \), from \( x = -2 \) to \( x = 1 \). The area \( A \) is given by:
\[
A = \int_{-2}^{1} \left[\text{Upper function} - \text{Lower function}\right] \, dx
\]
Here, the upper function is \( y = 3 \) and the lower function is \( y = \sqrt{x} + 2 \).

Thus, the area is:
\[
A = \int_{-2}^{1} \left(3 - (\sqrt{x} + 2)\right) \, dx = \int_{-2}^{1} \left(1 - \sqrt{x}\right) \, dx
\]

### Step 3: Compute the Integral
Let's calculate the integral:
\[
A = \int_{-2}^{1} \left(1 - \sqrt{x}\right) \, dx
\]

This can be split into two integrals:
\[
A = \int_{-2}^{1} 1 \, dx - \int_{-2}^{1} \sqrt{x} \, dx
\]

#### Integral 1:
\[
\int_{-2}^{1} 1 \, dx = x \bigg|_{-2}^{1} = 1 - (-2) = 3
\]

#### Integral 2:
For \( \int_{-2}^{1} \sqrt{x} \, dx \), this is:
\[
\int_{0}^{1} \sqrt{x} \, dx = \int_{0}^{1} x^{1/2} \, dx
\]
Using the power rule for integration:
\[
\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C
\]
We have:
\[
\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}
\]
Evaluate it from \( x = 0 \) to \( x = 1 \):
\[
\left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3}
\]

### Step 4: Combine Results
Thus, the area \( A \) is:
\[
A = 3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3} \approx 2.33
\]

So, the area of the region is approximately \( \boxed{2.33} \) square units.

Would you like more details or have any other questions?

Here are some related questions to explore:
1. What if the bounds of the integral were different?
2. How does changing the equation \( y = \sqrt{x} + 2 \) to \( y = \sqrt{x} + 3 \) affect the area?
3. Can this problem be solved by integrating with respect to \( y \)?
4. How would the area change if \( y = 3 \) was replaced by \( y = 4 \)?
5. What is the significance of the points of intersection in determining the bounds of the integral?

**Tip:** When calculating areas between curves, carefully identify the upper and lower functions within the interval of integration to avoid mistakes.

Use integration to find the area of the region enclosed by the graphs of y = √x + 2, x = -2, and y = 3. Round your result to 2 decimal places.

Discover how to calculate the area of the region enclosed by the curves y = √x + 2, x = -2, and y = 3 using definite integration. This detailed solution covers concepts like definite integrals, power rule for integration, and fundamental calculus theorems. Ideal for Grades 11-12 and college-level students.

Math Problem Statement

Use integration to find the area of the region enclosed by the graphs of y

√ x + 2 , x

− 2 , and y

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation