calculate the area of the regind bounded by
and -axis from to

Ask by Deleon Hobbs. in Ethiopia

Jan 02,2025

Question

calculate the area of the regind bounded by $$ F(x)=x^{2}-3 x+2 $$ and $$ x $$-axis from $$ x=0 $$ to $$ x-3 $$

UpStudy AI · Accepted Answer

To calculate the area of the region bounded by the function $$ F(x) = x^2 - 3x + 2 $$ and the $$ x $$-axis from $$ x = 0 $$ to $$ x = 3 $$, follow these steps:

### 1. **Find the Points of Intersection with the $$ x $$-Axis**
Set $$ F(x) = 0 $$:

$$
x^2 - 3x + 2 = 0
$$

Factor the quadratic equation:

$$
(x - 1)(x - 2) = 0
$$

So, the roots are $$ x = 1 $$ and $$ x = 2 $$. These are the points where the function intersects the $$ x $$-axis.

### 2. **Determine Where $$ F(x) $$ is Above or Below the $$ x $$-Axis**
- **From $$ x = 0 $$ to $$ x = 1 $$:**  
  Pick $$ x = 0.5 $$:  
  $$ F(0.5) = (0.5)^2 - 3(0.5) + 2 = 0.25 - 1.5 + 2 = 0.75 $$ (Positive)

- **From $$ x = 1 $$ to $$ x = 2 $$:**  
  Pick $$ x = 1.5 $$:  
  $$ F(1.5) = (1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 $$ (Negative)

- **From $$ x = 2 $$ to $$ x = 3 $$:**  
  Pick $$ x = 2.5 $$:  
  $$ F(2.5) = (2.5)^2 - 3(2.5) + 2 = 6.25 - 7.5 + 2 = 0.75 $$ (Positive)

### 3. **Set Up the Integrals for Each Region**

Since the function changes sign between $$ x = 1 $$ and $$ x = 2 $$, we'll compute the area in three parts:

$$
\text{Area} = \int_{0}^{1} F(x) \, dx + \int_{1}^{2} (-F(x)) \, dx + \int_{2}^{3} F(x) \, dx
$$

### 4. **Compute the Integrals**

First, find the indefinite integral of $$ F(x) $$:

$$
\int F(x) \, dx = \int (x^2 - 3x + 2) \, dx = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x
$$

Now, evaluate each definite integral:

- **From $$ x = 0 $$ to $$ x = 1 $$:**

$$
\left[ \frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x \right]_0^1 = \left( \frac{1}{3} - \frac{3}{2} + 2 \right) - 0 = \frac{5}{6}
$$

- **From $$ x = 1 $$ to $$ x = 2 $$:**

$$
\left[ \frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x \right]_1^2 = \left( \frac{8}{3} - 6 + 4 \right) - \left( \frac{1}{3} - \frac{3}{2} + 2 \right) = -\frac{1}{6}
$$

Since $$ F(x) $$ is negative in this interval, we take the absolute value:

$$
\int_{1}^{2} (-F(x)) \, dx = \frac{1}{6}
$$

- **From $$ x = 2 $$ to $$ x = 3 $$:**

$$
\left[ \frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x \right]_2^3 = \left( \frac{27}{3} - \frac{27}{2} + 6 \right) - \left( \frac{8}{3} - 6 + 4 \right) = \frac{5}{6}
$$

### 5. **Sum the Areas**

$$
\text{Total Area} = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6} \text{ square units}
$$

### **Final Answer**

The area of the region is $$ \frac{11}{6} $$ square units.
The area of the region is $$ \frac{11}{6} $$ square units.

calculate the area of the regind bounded by
and -axis from to

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