`int_(0)^(2)e^(x)\ dx`

Explanation

To solve the integral \( \int_{0}^{2} e^{x} \, dx \) using the limit of sums, we can follow these steps:

### Step 1: Define the Integral as a Limit of Sums
The integral from \( a \) to \( b \) of a function \( f(x) \) can be expressed as:
\[ \int_{a}^{b} f(x) \, dx = \lim_{h \to 0} \sum_{i=0}^{n-1} f(a + ih) \cdot h \]
where \( n = \frac{b-a}{h} \).

### Step 2: Set Up the Parameters
For our integral \( \int_{0}^{2} e^{x} \, dx \):
- \( a = 0 \)
- \( b = 2 \)
- Therefore, \( n = \frac{2-0}{h} = \frac{2}{h} \)

### Step 3: Write the Function in the Limit of Sums
The function \( f(x) = e^{x} \). Thus, we can write:
\[ \int_{0}^{2} e^{x} \, dx = \lim_{h \to 0} \sum_{i=0}^{n-1} e^{0 + ih} \cdot h = \lim_{h \to 0} \sum_{i=0}^{\frac{2}{h}-1} e^{ih} \cdot h \]

### Step 4: Evaluate the Sum
The sum can be expressed as:
\[ \sum_{i=0}^{\frac{2}{h}-1} e^{ih} = e^{0} + e^{h} + e^{2h} + \ldots + e^{\left(\frac{2}{h}-1\right)h} \]
This is a geometric series with:
- First term \( a = 1 \) (since \( e^{0} = 1 \))
- Common ratio \( r = e^{h} \)
- Number of terms \( n = \frac{2}{h} \)

The sum of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting our values:
\[ S = 1 \cdot \frac{(e^{h})^{\frac{2}{h}} - 1}{e^{h} - 1} = \frac{e^{2} - 1}{e^{h} - 1} \]

### Step 5: Substitute Back into the Limit
Now substituting back into the limit:
\[ \int_{0}^{2} e^{x} \, dx = \lim_{h \to 0} h \cdot \frac{e^{2} - 1}{e^{h} - 1} \]

### Step 6: Simplify Using the Limit Property
Using the property that \( \lim_{h \to 0} \frac{h}{e^{h} - 1} = 1 \):
\[ \int_{0}^{2} e^{x} \, dx = (e^{2} - 1) \cdot \lim_{h \to 0} \frac{h}{e^{h} - 1} = e^{2} - 1 \]

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

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