The graph of the function $g(x) = e^x$, $x \in \mathbb{R}$, $0 \leq x \leq 1$, is shown on the diagram below - Leaving Cert Mathematics - Question 6 - 2017

To find the area enclosed, we can set up the integral of the difference between the two functions from $0$ to $0.75$:

$$A = \int_{0}^{0.75} (g(x) - h(x)) \, dx = \int_{0}^{0.75} (e^x - e^{-x}) \, dx.$$

Calculating the antiderivatives:

$$\int e^x \, dx = e^x + C, \quad \int e^{-x} \, dx = -e^{-x} + C.$$

Thus,

$$A = \left[ e^{x} + e^{-x} \right]_{0}^{0.75} = (e^{0.75} + e^{-0.75}) - (e^{0} + e^{-0}).$$

Calculating this gives:

A = (e^{0.75} + e^{-0.75}) - (1 + 1) = e^{0.75} + e^{-0.75} - 2.\

Using a calculator,

$A \approx 0.5894.$

Thus, the area enclosed is approximately $0.5894$.