Evaluate the following integrals as limit of sums:
`int_(0)^(4)(3x^(2)+e^(2x))dx`

Answer

To evaluate the integral \(\int_{0}^{4} (3x^{2} + e^{2x}) \, dx\) using the limit of sums, we will follow these steps:

### Step 1: Define the function and the interval
We have the function \(f(x) = 3x^2 + e^{2x}\) and we want to evaluate the integral from \(a = 0\) to \(b = 4\).

### Step 2: Determine the width of each subinterval
We divide the interval \([0, 4]\) into \(n\) equal parts. The width of each subinterval is given by:
\[ h = \frac{b - a}{n} = \frac{4 - 0}{n} = \frac{4}{n} \]

### Step 3: Write the sum
The Riemann sum for the integral can be expressed as:
\[ \sum_{i=0}^{n-1} f(a + ih) \cdot h = \sum_{i=0}^{n-1} f\left(0 + i \cdot \frac{4}{n}\right) \cdot \frac{4}{n} \]
This simplifies to:
\[ \sum_{i=0}^{n-1} f\left(\frac{4i}{n}\right) \cdot \frac{4}{n} \]

### Step 4: Substitute the function into the sum
Now, we substitute \(f\left(\frac{4i}{n}\right)\):
\[ f\left(\frac{4i}{n}\right) = 3\left(\frac{4i}{n}\right)^2 + e^{2\left(\frac{4i}{n}\right)} = 3\frac{16i^2}{n^2} + e^{\frac{8i}{n}} \]
Thus, the sum becomes:
\[ \sum_{i=0}^{n-1} \left(3\frac{16i^2}{n^2} + e^{\frac{8i}{n}}\right) \cdot \frac{4}{n} \]
This can be separated into two sums:
\[ \frac{4}{n} \sum_{i=0}^{n-1} 3\frac{16i^2}{n^2} + \frac{4}{n} \sum_{i=0}^{n-1} e^{\frac{8i}{n}} \]

### Step 5: Simplify the sums
The first sum is:
\[ \frac{4 \cdot 48}{n^3} \sum_{i=0}^{n-1} i^2 = \frac{192}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} \approx \frac{192n^3}{6n^3} = 32 \quad \text{(as \(n \to \infty\))} \]

The second sum is a geometric series:
\[ \frac{4}{n} \sum_{i=0}^{n-1} e^{\frac{8i}{n}} \approx \frac{4}{n} \cdot \frac{1 - e^{8}}{1 - e^{\frac{8}{n}}} \approx 4 \cdot \frac{1 - e^{8}}{8} = \frac{4(1 - e^{8})}{8} = \frac{1 - e^{8}}{2} \]

### Step 6: Combine the results
Combining both parts, we get:
\[ \int_{0}^{4} (3x^{2} + e^{2x}) \, dx = 32 + \frac{1 - e^{8}}{2} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{4} (3x^{2} + e^{2x}) \, dx = 32 + \frac{1 - e^{8}}{2} \]