Compute the exact value of the area of the region formed between the graphs of f(x)=x^2-11x+27 and g(x)=x-5

To find the exact area of the region between the curves of \( f(x) = x^2 - 11x + 27 \) and \( g(x) = x - 5 \), we follow these steps:

### 1. Find the points of intersection.
Set \( f(x) = g(x) \) to find where the curves intersect.

\[
x^2 - 11x + 27 = x - 5
\]

Rearranging the equation:

\[
x^2 - 11x + 27 - x + 5 = 0
\]

\[
x^2 - 12x + 32 = 0
\]

Now, solve the quadratic equation:

\[
x^2 - 12x + 32 = 0
\]

Using the quadratic formula:

\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)}
\]

\[
x = \frac{12 \pm \sqrt{144 - 128}}{2}
\]

\[
x = \frac{12 \pm \sqrt{16}}{2}
\]

\[
x = \frac{12 \pm 4}{2}
\]

Thus, the solutions are:

\[
x = \frac{12 + 4}{2} = 8 \quad \text{and} \quad x = \frac{12 - 4}{2} = 4
\]

So, the points of intersection are \( x = 4 \) and \( x = 8 \).

### 2. Set up the integral for the area.
The area between the two curves from \( x = 4 \) to \( x = 8 \) is given by the integral of the difference between \( g(x) \) and \( f(x) \):

\[
\text{Area} = \int_{4}^{8} (g(x) - f(x)) \, dx
\]

Substitute \( g(x) = x - 5 \) and \( f(x) = x^2 - 11x + 27 \):

\[
\text{Area} = \int_{4}^{8} \left[(x - 5) - (x^2 - 11x + 27)\right] \, dx
\]

Simplify the integrand:

\[
\text{Area} = \int_{4}^{8} \left(x - 5 - x^2 + 11x - 27\right) \, dx
\]

\[
\text{Area} = \int_{4}^{8} \left(-x^2 + 12x - 32\right) \, dx
\]

### 3. Compute the integral.
Now, integrate term by term:

\[
\int \left(-x^2 + 12x - 32\right) \, dx = -\frac{x^3}{3} + 6x^2 - 32x
\]

Evaluate the definite integral from \( x = 4 \) to \( x = 8 \):

\[
\left[-\frac{x^3}{3} + 6x^2 - 32x\right]_{4}^{8}
\]

First, evaluate at \( x = 8 \):

\[
-\frac{8^3}{3} + 6(8^2) - 32(8) = -\frac{512}{3} + 6(64) - 256 = -\frac{512}{3} + 384 - 256 = -\frac{512}{3} + 128
\]

Now, evaluate at \( x = 4 \):

\[
-\frac{4^3}{3} + 6(4^2) - 32(4) = -\frac{64}{3} + 6(16) - 128 = -\frac{64}{3} + 96 - 128 = -\frac{64}{3} - 32
\]

Now, subtract the two results:

\[
\left(-\frac{512}{3} + 128\right) - \left(-\frac{64}{3} - 32\right) = -\frac{512}{3} + 128 + \frac{64}{3} + 32
\]

Simplify:

\[
-\frac{512}{3} + \frac{64}{3} = -\frac{448}{3}
\]

\[
128 + 32 = 160
\]

So, the total area is:

\[
-\frac{448}{3} + 160 = \frac{-448 + 480}{3} = \frac{32}{3}
\]

### 4. Conclusion
The exact value of the area of the region is \( \frac{32}{3} \) square units.

Do you need further clarification or details? Here are 5 related questions:

1. How do you compute the points of intersection algebraically for two curves?
2. Why is the area between two curves expressed as the integral of their difference?
3. How do you set up an integral for the area when curves cross multiple times?
4. What are the geometric interpretations of definite integrals?
5. Can you apply numerical methods to approximate the area between curves?

**Tip:** Always check the order of the curves when integrating for area, as subtracting the wrong way might lead to negative areas.

Learn how to compute the exact area of the region between the curves f(x)=x^2-11x+27 and g(x)=x-5. This step-by-step solution covers key calculus concepts, including definite integrals, quadratic equations, and algebra. Suitable for advanced high school students in Grades 11-12.

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