Evaluate: `int_-1^2 (e^(3x)+7x-5) dx` as a limit of sums.

Answer

To evaluate the integral \(\int_{-1}^{2} (e^{3x} + 7x - 5) \, dx\) as a limit of sums, we will follow the steps outlined below:

### Step 1: Define the function and interval
Let \( f(x) = e^{3x} + 7x - 5 \). We are integrating this function over the interval \([-1, 2]\).

### Step 2: Determine the width of subintervals
The width of each subinterval, denoted as \( h \), can be calculated as:
\[ h = \frac{b - a}{n} = \frac{2 - (-1)}{n} = \frac{3}{n} \]

### Step 3: Set up the limit of sums
The Riemann sum for the integral can be expressed as:
\[ \int_{-1}^{2} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(a + ih) \cdot h \]
Substituting \( a = -1 \) and \( h = \frac{3}{n} \):
\[ \int_{-1}^{2} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(-1 + i \cdot \frac{3}{n}\right) \cdot \frac{3}{n} \]

### Step 4: Substitute \( f(x) \) into the sum
Now we need to evaluate \( f\left(-1 + i \cdot \frac{3}{n}\right) \):
\[ f\left(-1 + i \cdot \frac{3}{n}\right) = e^{3\left(-1 + i \cdot \frac{3}{n}\right)} + 7\left(-1 + i \cdot \frac{3}{n}\right) - 5 \]
This simplifies to:
\[ = e^{-3} \cdot e^{\frac{9i}{n}} + 7\left(-1 + \frac{3i}{n}\right) - 5 \]
\[ = e^{-3} \cdot e^{\frac{9i}{n}} - 7 + \frac{21i}{n} - 5 \]
\[ = e^{-3} \cdot e^{\frac{9i}{n}} - 12 + \frac{21i}{n} \]

### Step 5: Substitute back into the Riemann sum
Now, substituting this back into the sum:
\[ \int_{-1}^{2} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left(e^{-3} \cdot e^{\frac{9i}{n}} - 12 + \frac{21i}{n}\right) \cdot \frac{3}{n} \]
Distributing \( \frac{3}{n} \):
\[ = \lim_{n \to \infty} \left( \sum_{i=0}^{n-1} \left(e^{-3} \cdot e^{\frac{9i}{n}} \cdot \frac{3}{n}\right) - 12 \cdot \frac{3}{n} \sum_{i=0}^{n-1} 1 + \frac{21}{n} \sum_{i=0}^{n-1} i \right) \]

### Step 6: Evaluate the sums
1. The first sum:
\[ \sum_{i=0}^{n-1} e^{-3} \cdot e^{\frac{9i}{n}} \cdot \frac{3}{n} \to 3 e^{-3} \int_{0}^{1} e^{9x} \, dx = 3 e^{-3} \cdot \left[\frac{1}{9} e^{9x}\right]_{0}^{1} = \frac{3 e^{-3}}{9} (e^9 - 1) \]

2. The second sum:
\[ -12 \cdot \frac{3}{n} \cdot n = -36 \]

3. The third sum:
\[ \frac{21}{n} \cdot \frac{(n-1)n}{2} \to \frac{21}{2} \cdot n \to \frac{21}{2} \]

### Step 7: Combine the results
Putting it all together:
\[ \int_{-1}^{2} f(x) \, dx = \frac{3 e^{-3}}{9} (e^9 - 1) - 36 + \frac{21}{2} \]

### Step 8: Final simplification
Combine the constants and simplify:
\[ = \frac{e^{-3}}{3} (e^9 - 1) - 36 + 10.5 \]
\[ = \frac{e^{-3}}{3} (e^9 - 1) - 25.5 \]

### Conclusion
Thus, the evaluation of the integral \(\int_{-1}^{2} (e^{3x} + 7x - 5) \, dx\) as a limit of sums leads to the final expression.

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