`int_0^2e^xdx` as the limit of a sum

Answer

To evaluate the integral \( \int_0^2 e^x \, dx \) as the limit of a sum, we can follow these steps:

### Step 1: Define the Integral
We start with the integral we want to evaluate:
\[ I = \int_0^2 e^x \, dx \]

### Step 2: Set Up the Limit of a Sum
The integral can be expressed as a limit of a Riemann sum. We divide the interval \([0, 2]\) into \(n\) equal subintervals. The width of each subinterval is:
\[ h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n} \]

### Step 3: Write the Riemann Sum
The Riemann sum can be expressed as:
\[ I \approx h \sum_{r=0}^{n-1} f(a + rh) = h \sum_{r=0}^{n-1} e^{0 + rh} = h \sum_{r=0}^{n-1} e^{rh} \]
Substituting \(h\):
\[ I = \frac{2}{n} \sum_{r=0}^{n-1} e^{\frac{2r}{n}} \]

### Step 4: Take the Limit as \(n \to \infty\)
Now, we express the integral as the limit of the sum:
\[ I = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} e^{\frac{2r}{n}} \]

### Step 5: Recognize the Sum as a Riemann Sum
As \(n\) approaches infinity, the sum approaches the integral:
\[ I = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} e^{\frac{2r}{n}} \to \int_0^2 e^x \, dx \]

### Step 6: Evaluate the Integral
Now we can evaluate the integral:
\[ I = \left[ e^x \right]_0^2 = e^2 - e^0 = e^2 - 1 \]

### Final Result
Thus, the value of the integral is:
\[ I = e^2 - 1 \]