Evaluate the following integrals as limit of sums:
`int_(0)^(1)e^(2-3x)dx`

Answer

To evaluate the integral \( \int_{0}^{1} e^{2 - 3x} \, dx \) using the limit of sums, we will follow these steps:

### Step 1: Define the integral as a limit of sums
The integral can be expressed as:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(x_i) \Delta x \]
where \( \Delta x = \frac{b-a}{n} \) and \( x_i = a + i \Delta x \).

### Step 2: Identify \( a \), \( b \), and \( f(x) \)
For our integral:
- \( a = 0 \)
- \( b = 1 \)
- \( f(x) = e^{2 - 3x} \)

### Step 3: Calculate \( \Delta x \)
\[ \Delta x = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n} \]

### Step 4: Express \( x_i \)
The points \( x_i \) can be expressed as:
\[ x_i = 0 + i \Delta x = \frac{i}{n} \]

### Step 5: Substitute into the sum
Now, we can write the limit of sums:
\[ \int_{0}^{1} e^{2 - 3x} \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} e^{2 - 3\left(\frac{i}{n}\right)} \cdot \frac{1}{n} \]
This simplifies to:
\[ = \lim_{n \to \infty} \frac{1}{n} \sum_{i=0}^{n-1} e^{2 - \frac{3i}{n}} \]

### Step 6: Factor out constants
Factor out the constant \( e^2 \):
\[ = e^2 \lim_{n \to \infty} \frac{1}{n} \sum_{i=0}^{n-1} e^{-\frac{3i}{n}} \]

### Step 7: Recognize the sum as a Riemann sum
The sum \( \frac{1}{n} \sum_{i=0}^{n-1} e^{-\frac{3i}{n}} \) approaches the integral \( \int_{0}^{1} e^{-3x} \, dx \) as \( n \to \infty \).

### Step 8: Evaluate the integral
Now we need to evaluate:
\[ \int_{0}^{1} e^{-3x} \, dx \]
The antiderivative of \( e^{-3x} \) is:
\[ -\frac{1}{3} e^{-3x} \]
Thus,
\[ \int_{0}^{1} e^{-3x} \, dx = \left[-\frac{1}{3} e^{-3x}\right]_{0}^{1} = -\frac{1}{3} (e^{-3} - e^{0}) = -\frac{1}{3} (e^{-3} - 1) = \frac{1 - e^{-3}}{3} \]

### Step 9: Combine results
Now substituting back into our expression:
\[ \int_{0}^{1} e^{2 - 3x} \, dx = e^2 \cdot \frac{1 - e^{-3}}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} e^{2 - 3x} \, dx = \frac{e^2 (1 - e^{-3})}{3} \]