Question content area top

Part 1

Find the area of the region described.

The region bounded by

yequals=StartAbsoluteValue x minus 3 EndAbsoluteValuex−3

and

yequals=StartFraction x Over 2 EndFractionx2

Question content area bottom

Part 1

The area of the region bounded by

yequals=StartAbsoluteValue x minus 3 EndAbsoluteValuex−3

and

yequals=StartFraction x Over 2 EndFractionx2

is

enter your response here.

​(Type an exact​ answer, using radicals as​ needed.)

To find the area of the region bounded by the two functions \( y = |x - 3| \) and \( y = \frac{x}{2} \), we need to:

1. Find the points of intersection between the two functions.
2. Set up the integral(s) to calculate the area between these curves.
3. Evaluate the integral(s).

### Step 1: Find the Points of Intersection

The points of intersection occur where the two functions are equal:

\[
|x - 3| = \frac{x}{2}
\]

This equation needs to be split into two cases, based on the definition of the absolute value.

**Case 1: \( x - 3 \geq 0 \) (i.e., \( x \geq 3 \))**

\[
x - 3 = \frac{x}{2}
\]

Solve for \( x \):

\[
2(x - 3) = x
\]

\[
2x - 6 = x
\]

\[
x = 6
\]

**Case 2: \( x - 3 < 0 \) (i.e., \( x < 3 \))**

\[
-(x - 3) = \frac{x}{2}
\]

\[
-2(x - 3) = x
\]

\[
-2x + 6 = x
\]

\[
3x = 6
\]

\[
x = 2
\]

So, the points of intersection are \( x = 2 \) and \( x = 6 \).

### Step 2: Set Up the Integral

The area between the curves is given by the integral of the difference between the upper and lower functions, from \( x = 2 \) to \( x = 6 \).

\[
A = \int_{2}^{6} \left[ |x - 3| - \frac{x}{2} \right] \, dx
\]

We need to evaluate this integral in two parts, considering the definition of \( |x - 3| \).

- From \( x = 2 \) to \( x = 3 \), \( |x - 3| = -(x - 3) = -x + 3 \).
- From \( x = 3 \) to \( x = 6 \), \( |x - 3| = x - 3 \).

So, we split the integral:

\[
A = \int_{2}^{3} \left[ (-x + 3) - \frac{x}{2} \right] \, dx + \int_{3}^{6} \left[ (x - 3) - \frac{x}{2} \right] \, dx
\]

### Step 3: Evaluate the Integral

#### First Integral: \( \int_{2}^{3} \left[ -x + 3 - \frac{x}{2} \right] \, dx \)

\[
= \int_{2}^{3} \left[ -\frac{3x}{2} + 3 \right] \, dx
\]

\[
= \left[ -\frac{3}{4}x^2 + 3x \right]_{2}^{3}
\]

\[
= \left(-\frac{3}{4}(9) + 9 \right) - \left(-\frac{3}{4}(4) + 6 \right)
\]

\[
= \left(-\frac{27}{4} + 9 \right) - \left(-3 + 6 \right)
\]

\[
= \left(-\frac{27}{4} + \frac{36}{4} \right) - 3 + 6
\]

\[
= \frac{9}{4} + 3 = \frac{21}{4}
\]

#### Second Integral: \( \int_{3}^{6} \left[ x - 3 - \frac{x}{2} \right] \, dx \)

\[
= \int_{3}^{6} \left[ \frac{x}{2} - 3 \right] \, dx
\]

\[
= \left[ \frac{1}{4}x^2 - 3x \right]_{3}^{6}
\]

\[
= \left(\frac{1}{4}(36) - 18 \right) - \left(\frac{1}{4}(9) - 9 \right)
\]

\[
= \left(9 - 18 \right) - \left(\frac{9}{4} - 9 \right)
\]

\[
= -9 - \frac{27}{4} + \frac{36}{4} 
\]

\[
= -9 + \frac{9}{4} = \frac{-27}{4}
\]

### Combine the Results

\[
A = \frac{21}{4} + \frac{27}{4} = \frac{48}{4} = 12
\]

So, the exact area of the region is \( \boxed{12} \).

Do you have any questions or need further details on any of these steps?

---

**5 Related Questions:**

1. How would you find the area if the functions were different?
2. What happens if one of the functions is non-linear, such as a quadratic function?
3. How would the procedure change if the bounds were not the x-axis but y-axis?
4. Can you find the area between two curves using polar coordinates?
5. How can you determine the intersection points for more complex functions?

**Tip:** Always check the domain where the functions intersect to ensure the correct setup of integrals.

Discover how to find the area of the region bounded by the absolute value function |x - 3| and the linear function x/2. This problem involves setting up integrals and evaluating them step-by-step. Suitable for advanced high school students.

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