Calculate integral from 0 to 1 e^x dx using the limit

Calculate $\int_0^1 e^x \, dx$ using the limit of Riemann sums.

We will evaluate the integral using the definition as the limit of Riemann sums.

Step 1: Partition the interval

Divide the interval $[0,1]$ into $n$ equal subintervals. The width of each subinterval is:

\[\Delta x = \frac{1-0}{n} = \frac{1}{n}\]

The partition points are:

\[x_i = 0 + i \cdot \Delta x = \frac{i}{n}, \quad i = 0, 1, 2, \ldots, n\]

Step 2: Choose the sample points

We use the right endpoint of each subinterval:

\[\xi_i = x_i = \frac{i}{n}\]

Step 3: Form the Riemann sum

The Riemann sum is:

\[S_n = \sum_{i=1}^{n} f(\xi_i) \Delta x = \sum_{i=1}^{n} e^{i/n} \cdot \frac{1}{n} = \frac{1}{n} \sum_{i=1}^{n} e^{i/n}\]

Step 4: Simplify the sum

\[S_n = \frac{1}{n} \sum_{i=1}^{n} e^{i/n} = \frac{1}{n} \sum_{i=1}^{n} \left(e^{1/n}\right)^i\]

This is a geometric series with first term $a = e^{1/n}$ and common ratio $r = e^{1/n}$:

\[\sum_{i=1}^{n} \left(e^{1/n}\right)^i = e^{1/n} \cdot \frac{1 - \left(e^{1/n}\right)^n}{1 - e^{1/n}} = e^{1/n} \cdot \frac{1 - e}{1 - e^{1/n}}\]

Therefore:

\[S_n = \frac{1}{n} \cdot e^{1/n} \cdot \frac{1 - e}{1 - e^{1/n}} = \frac{e^{1/n}(1 - e)}{n(1 - e^{1/n})}\]

Step 5: Take the limit as $n \to \infty$

We rewrite:

\[S_n = \frac{e^{1/n} - e^{1+1/n}}{n(1 - e^{1/n})} = \frac{e^{1/n} - e \cdot e^{1/n}}{n(1 - e^{1/n})} = \frac{e^{1/n}(1-e)}{n(1-e^{1/n})}\]

Multiply numerator and denominator by $-1$:

\[S_n = \frac{e^{1/n}(e-1)}{n(e^{1/n}-1)}\]

As $n \to \infty$, let $h = \frac{1}{n} \to 0$:

\[\lim_{n \to \infty} S_n = \lim_{h \to 0} \frac{e^h(e-1)}{(e^h-1)/h} = (e-1) \cdot \lim_{h \to 0} \frac{h \cdot e^h}{e^h-1}\]

Since $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$, we have $\lim_{h \to 0} \frac{h}{e^h-1} = 1$:

\[\lim_{n \to \infty} S_n = (e-1) \cdot e^0 \cdot 1 = e - 1\]

Verification using antiderivative:

\[\int_0^1 e^x \, dx = \left[e^x\right]_0^1 = e^1 - e^0 = e - 1\]

Final answer:

\[\boxed{\int_0^1 e^x \, dx = e - 1}\]