Find the area of the region bounded by `y=3^(x)` and the lines y = 3 and x = 0.

Answer

To find the area of the region bounded by the curve \( y = 3^x \), the line \( y = 3 \), and the line \( x = 0 \), we can follow these steps:

### Step 1: Determine the points of intersection
First, we need to find the points where the curve \( y = 3^x \) intersects the line \( y = 3 \). We set the equations equal to each other:

\[ 3^x = 3 \]

Taking the logarithm (base 3) of both sides, we have:

\[ x = 1 \]

Thus, the points of intersection are \( (1, 3) \) and \( (0, 1) \) when \( x = 0 \).

### Step 2: Set up the integral for the area
The area \( A \) between the curve and the line can be found using the integral:

\[ A = \int_{0}^{1} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]

Here, \( y_{\text{top}} = 3 \) (the line) and \( y_{\text{bottom}} = 3^x \) (the curve). Therefore, we can write:

\[ A = \int_{0}^{1} (3 - 3^x) \, dx \]

### Step 3: Evaluate the integral
Now we can evaluate the integral:

\[ A = \int_{0}^{1} 3 \, dx - \int_{0}^{1} 3^x \, dx \]

Calculating the first integral:

\[ \int_{0}^{1} 3 \, dx = 3[x]_{0}^{1} = 3(1 - 0) = 3 \]

Next, we calculate the second integral. The integral of \( 3^x \) can be computed as follows:

\[ \int 3^x \, dx = \frac{3^x}{\ln(3)} + C \]

Thus,

\[ \int_{0}^{1} 3^x \, dx = \left[ \frac{3^x}{\ln(3)} \right]_{0}^{1} = \frac{3^1}{\ln(3)} - \frac{3^0}{\ln(3)} = \frac{3}{\ln(3)} - \frac{1}{\ln(3)} = \frac{3 - 1}{\ln(3)} = \frac{2}{\ln(3)} \]

### Step 4: Combine the results
Now we can combine the results of the two integrals:

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

### Final Answer
Thus, the area of the region bounded by the curve \( y = 3^x \), the line \( y = 3 \), and the line \( x = 0 \) is:

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

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