Questions: Find the area between the curves. x=-2, x=4, y=3e^(3x), y=2e^(3x)+1 The area between the curves is approximately.

Solution Steps

To find the area between the curves \( y = 3e^{3x} \) and \( y = 2e^{3x} + 1 \) from \( x = -2 \) to \( x = 4 \), we first express the area \( A \) as the integral of the difference between the two functions over the specified interval:

\[ A = \int_{-2}^{4} \left( 3e^{3x} - (2e^{3x} + 1) \right) \, dx \]

We simplify the integrand:

\[ 3e^{3x} - (2e^{3x} + 1) = 3e^{3x} - 2e^{3x} - 1 = e^{3x} - 1 \]

Thus, the area can be rewritten as:

\[ A = \int_{-2}^{4} (e^{3x} - 1) \, dx \]

Now we compute the integral:

\[ A = \int_{-2}^{4} e^{3x} \, dx - \int_{-2}^{4} 1 \, dx \]

Calculating each part separately, we find:

1. The integral of \( e^{3x} \):

\[ \int e^{3x} \, dx = \frac{1}{3} e^{3x} \]

Evaluating from \( -2 \) to \( 4 \):

\[ \left[ \frac{1}{3} e^{3(4)} - \frac{1}{3} e^{3(-2)} \right] \]

2. The integral of \( 1 \):

\[ \int 1 \, dx = x \]

Evaluating from \( -2 \) to \( 4 \):

\[ 4 - (-2) = 6 \]

Combining these results gives us the total area \( A \).

Final Answer