Evaluate $\int\limits_{0}^{2} e^x dx$

$e^2-1$

The correct answer is Option (3) → $e^2-1$

We know that $\int\limits_{a}^{b} f(x) dx = \lim\limits_{n \to \infty} h \sum\limits_{r=0}^{n-1} f(a + rh)$

For $I = \int\limits_{0}^{2} e^x dx$, we have $a = 0$ and $b = 2$

$∴h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n} ⇒nh = 2$

$∴I = \int\limits_{0}^{2} e^x dx$

$= \lim\limits_{h \to 0} h [1 + e^h + e^{2h} + \dots + e^{(n-1)h}]$

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

$= \lim\limits_{h \to 0} h \left( \frac{e^{nh} - 1}{e^h - 1} \right) = \lim\limits_{h \to 0} h \left( \frac{e^2 - 1}{e^h - 1} \right)$

$= (e^2 - 1) \cdot \lim\limits_{h \to 0} \frac{h}{e^h - 1}$

$= e^2 – 1$