Evaluate $\int\limits_{0}^{2} e^x \, dx$ as the limit of a sum.

$e^2 - 1$

The correct answer is Option (4) → $e^2 - 1$

By definition

$\int\limits_{0}^{2} e^x \, dx = (2 - 0) \lim\limits_{n \to \infty} \frac{1}{n} \left[ e^0 + e^{\frac{2}{n}} + e^{\frac{4}{n}} + \dots + e^{\frac{2n-2}{n}} \right] \text{}$

Using the sum to $n$ terms of a G.P., where $a = 1, r = e^{\frac{2}{n}}$, we have

$\int\limits_{0}^{2} e^x \, dx = 2 \lim\limits_{n \to \infty} \frac{1}{n} \left[ \frac{(e^{\frac{2}{n}})^n - 1}{e^{\frac{2}{n}} - 1} \right] = 2 \lim\limits_{n \to \infty} \frac{1}{n} \left[ \frac{e^2 - 1}{e^{\frac{2}{n}} - 1} \right]$

$= \frac{2(e^2 - 1)}{\lim\limits_{n \to \infty} \left[ \frac{e^{\frac{2}{n}} - 1}{\frac{2}{n}} \right] \cdot 2} = e^2 - 1 \qquad \left[ \text{using } \lim\limits_{h \to 0} \frac{e^h - 1}{h} = 1 \right]$