The diagram shows part of the graph of $y = a \cosh(bx)$ - Scottish Highers Maths - Question 12 - 2015

To find the value of (\int_0^{\frac{3\pi}{4}} \cosh(kx) , dx), we can use the antiderivative of (\cosh(kx)), which is (\frac{1}{k} \sinh(kx)). Therefore:

$$\int \cosh(kx) \, dx = \frac{1}{k} \sinh(kx) + C$$

Evaluating from (0) to (\frac{3\pi}{4}):

$$\int_0^{\frac{3\pi}{4}} \cosh(kx) \, dx = \left[ \frac{1}{k} \sinh(kx) \right]_0^{\frac{3\pi}{4}} = \frac{1}{k} \sinh\left(k \frac{3\pi}{4}\right) - \frac{1}{k} \sinh(0)$$

Observing that (\sinh(0) = 0), we have:

$$\int_0^{\frac{3\pi}{4}} \cosh(kx) \, dx = \frac{1}{k} \sinh\left(k \frac{3\pi}{4}\right)$$

Given that this area is (\frac{1}{2}) unit$^2$, we can equate:

$$\frac{1}{k} \sinh\left(k \frac{3\pi}{4}\right) = \frac{1}{2}$$

From this, we can solve for the specific values of (k).