R is the region bounded by the functions f(x)=2/x and g(x)=−x/6+13/6. Find the area A of R. Enter an exact answer.

To find the area \(A\) of the region \(R\) bounded by the functions \(f(x) = \frac{2}{x}\) and \(g(x) = -\frac{x}{6} + \frac{13}{6}\), we follow a similar process:

1. **Find the points of intersection** of \(f(x)\) and \(g(x)\).
2. **Set up the integral** representing the area between these curves over the interval determined by the intersection points.
3. **Evaluate the integral** to find the exact area.

### Step 1: Find the Points of Intersection

To find the points of intersection, set \(f(x) = g(x)\):

\[
\frac{2}{x} = -\frac{x}{6} + \frac{13}{6}
\]

Multiply both sides by \(6x\) to clear the fractions:

\[
6x \cdot \frac{2}{x} = 6x \left(-\frac{x}{6} + \frac{13}{6}\right)
\]

Simplify the equation:

\[
12 = -x^2 + 13x
\]

Rearrange it into a standard quadratic form:

\[
x^2 - 13x + 12 = 0
\]

Factor the quadratic equation:

\[
(x - 1)(x - 12) = 0
\]

So, the solutions are:

\[
x = 1 \quad \text{and} \quad x = 12
\]

These are the \(x\)-coordinates of the points where the curves intersect.

### Step 2: Set Up the Integral

Now we set up the integral to find the area between the curves from \(x = 1\) to \(x = 12\).

Since \(f(x) = \frac{2}{x}\) is above \(g(x) = -\frac{x}{6} + \frac{13}{6}\) in this interval, the area \(A\) is given by:

\[
A = \int_{1}^{12} \left( \frac{2}{x} - \left(-\frac{x}{6} + \frac{13}{6}\right) \right) dx
\]

Simplify the integrand:

\[
A = \int_{1}^{12} \left( \frac{2}{x} + \frac{x}{6} - \frac{13}{6} \right) dx
\]

### Step 3: Evaluate the Integral

Evaluate each term separately:

\[
\int \frac{2}{x} \, dx = 2 \ln|x|, \quad \int \frac{x}{6} \, dx = \frac{x^2}{12}, \quad \int \frac{13}{6} \, dx = \frac{13x}{6}
\]

Now compute the definite integral:

\[
A = \left[ 2 \ln|x| + \frac{x^2}{12} - \frac{13x}{6} \right]_{1}^{12}
\]

Evaluate at the upper limit \(x = 12\) and the lower limit \(x = 1\):

\[
A = \left(2 \ln 12 + \frac{12^2}{12} - \frac{13 \cdot 12}{6}\right) - \left(2 \ln 1 + \frac{1^2}{12} - \frac{13 \cdot 1}{6}\right)
\]

Simplify each part:

\[
A = \left(2 \ln 12 + 12 - 26\right) - \left(0 + \frac{1}{12} - \frac{13}{6}\right)
\]

Simplify further:

\[
A = (2 \ln 12 - 14) - \left(\frac{1}{12} - \frac{13}{6}\right)
\]

Combine like terms and simplify:

\[
A = 2 \ln 12 - 14 - \left(\frac{1}{12} - \frac{26}{12}\right)
\]

\[
A = 2 \ln 12 - 14 - \left(-\frac{25}{12}\right)
\]

\[
A = 2 \ln 12 - 14 + \frac{25}{12}
\]

Convert \(-14\) to a fraction with a denominator of 12:

\[
-14 = -\frac{168}{12}
\]

So the expression becomes:

\[
A = 2 \ln 12 - \frac{168}{12} + \frac{25}{12}
\]

\[
A = 2 \ln 12 - \frac{143}{12}
\]

### Conclusion

The exact area \(A\) of the region \(R\) is:

\[
A = 2 \ln 12 - \frac{143}{12}
\]

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How would the area calculation change if \(f(x)\) and \(g(x)\) were reversed in the interval?
2. What is the impact of different intersection points on the total area between two curves?
3. How can you use symmetry to simplify the calculation of areas between curves?
4. How would this integral change if one of the functions was a quadratic instead?
5. What is the significance of logarithmic terms in the calculation of areas for functions involving reciprocals?

**Tip:** Always check the graph of the functions to understand their intersection and relative positions before integrating.

Learn how to calculate the exact area of the region bounded by the curves f(x)=2/x and g(x)=-x/6+13/6 using integral calculus. Detailed step-by-step solution provided.

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