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?

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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.