Find the area bounded by the curves y=9x, y=2x-3, and x=1.

To find the area bounded by the curves \(y = 9x\), \(y = 2x - 3\), and the vertical line \(x = 1\), we need to identify the points of intersection between these curves and then set up integrals to compute the area.

### Step 1: Find the Points of Intersection

1. **Intersection of \(y = 9x\) and \(y = 2x - 3\):**

   Set \(9x = 2x - 3\):

   \[
   9x - 2x = -3
   \]

   \[
   7x = -3 \implies x = -\frac{3}{7}
   \]

   Now, find the corresponding \(y\)-coordinate:

   \[
   y = 9x = 9\left(-\frac{3}{7}\right) = -\frac{27}{7}
   \]

   So, the point of intersection is \(\left(-\frac{3}{7}, -\frac{27}{7}\right)\).

2. **Intersection of each line with \(x = 1\):**

   - For \(y = 9x\):

   \[
   y = 9(1) = 9
   \]

   So, the point is \((1, 9)\).

   - For \(y = 2x - 3\):

   \[
   y = 2(1) - 3 = -1
   \]

   So, the point is \((1, -1)\).

### Step 2: Determine the Boundaries of the Enclosed Region

The region is bounded between \(x = -\frac{3}{7}\) (where the two lines intersect) and \(x = 1\) (the vertical line).

- The curve \(y = 9x\) is above \(y = 2x - 3\) in this interval.

### Step 3: Set Up the Integral for the Area

The area \(A\) between \(x = -\frac{3}{7}\) and \(x = 1\) can be calculated by integrating the difference between the top function (\(y = 9x\)) and the bottom function (\(y = 2x - 3\)):

\[
A = \int_{-\frac{3}{7}}^{1} \left(9x - (2x - 3)\right) \, dx
\]

Simplify the integrand:

\[
A = \int_{-\frac{3}{7}}^{1} (9x - 2x + 3) \, dx = \int_{-\frac{3}{7}}^{1} (7x + 3) \, dx
\]

### Step 4: Compute the Integral

Evaluate the integral:

\[
A = \left[ \frac{7x^2}{2} + 3x \right]_{-\frac{3}{7}}^{1}
\]

Now, calculate the values at \(x = 1\) and \(x = -\frac{3}{7}\):

1. For \(x = 1\):

\[
\frac{7(1)^2}{2} + 3(1) = \frac{7}{2} + 3 = \frac{7}{2} + \frac{6}{2} = \frac{13}{2}
\]

2. For \(x = -\frac{3}{7}\):

\[
\frac{7\left(-\frac{3}{7}\right)^2}{2} + 3\left(-\frac{3}{7}\right)
\]

Compute each term separately:

\[
\frac{7\left(\frac{9}{49}\right)}{2} + 3\left(-\frac{3}{7}\right) = \frac{63}{98} - \frac{9}{7}
\]

Simplify:

\[
\frac{63}{98} = \frac{9}{14}, \quad \text{and} \quad -\frac{9}{7} = -\frac{18}{14}
\]

\[
\frac{9}{14} - \frac{18}{14} = -\frac{9}{14}
\]

### Step 5: Compute the Total Area

Now, compute the difference:

\[
A = \frac{13}{2} - \left(-\frac{9}{14}\right)
\]

Convert \(\frac{13}{2}\) to a fraction with a denominator of 14:

\[
\frac{13}{2} = \frac{91}{14}
\]

So:

\[
A = \frac{91}{14} + \frac{9}{14} = \frac{100}{14} = \frac{50}{7}
\]

### Conclusion

The area bounded by the curves \(y = 9x\), \(y = 2x - 3\), and \(x = 1\) is \(\frac{50}{7}\) square units.

Would you like any more details or have any questions?

Here are some related questions you might find interesting:

1. How do you find the area of a region bounded by multiple curves?
2. What is the importance of determining the points of intersection when calculating areas between curves?
3. Can you apply these methods to find volumes of solids of revolution?
4. How do the slopes of intersecting lines affect the area of the enclosed region?
5. What happens to the bounded area if the vertical line \(x = 1\) is moved further away?

**Tip:** Always sketch the curves and identify the intersection points clearly to avoid errors in setting up your integrals.

Learn how to calculate the area bounded by the curves y=9x, y=2x-3, and the line x=1 using calculus. This detailed solution includes finding intersection points and setting up integrals to compute the area. Suitable for advanced high school students and calculus enthusiasts.

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