The finite region $S$, shown shaded in Figure 2, is bounded by the $y$-axis, the $x$-axis, the line with equation $x = 4$ and the curve with equation $y = e^x + 2e^{-x}, \ x > 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 5

To find the volume of the solid generated when the region $S$ is rotated about the $x$-axis, we use the formula for the volume of revolution:

$V = \pi \int_a^b [f(x)]^2 \, dx$

where $f(x)$ represents the upper curve. Here, $f(x) = e^x + 2e^{-x}$, and the bounds of integration are from $0$ to $4$ (since $x = 4$).

Thus, we set up the integral:

$V = \pi \int_0^4 (e^x + 2e^{-x})^2 \, dx$

Now we expand the expression inside the integral:

$(e^x + 2e^{-x})^2 = e^{2x} + 4 + 4e^{-x} e^x = e^{2x} + 4 + 4$

So, the integral becomes:

$V = \pi \int_0^4 (e^{2x} + 4 + 4) \, dx$

We can now split the integral:

$V = \pi \left( \int_0^4 e^{2x} \, dx + \int_0^4 4 \, dx + \int_0^4 4e^{-x} \, dx \right)$

Next, we evaluate each integral separately:

1. For $\int_0^4 e^{2x} \, dx$, we have:

   $= \frac{1}{2} e^{2x} \bigg|_0^4 = \frac{1}{2} (e^8 - 1)$

2. For $\int_0^4 4 \, dx$:

   $= 4x \bigg|_0^4 = 16$

3. For $\int_0^4 4e^{-x} \, dx$:

   $= -4e^{-x} \bigg|_0^4 = -4(e^{-4} - 1) = 4 - 4e^{-4}$

Putting it all together, we have:

$V = \pi \left( \frac{1}{2} (e^8 - 1) + 16 + 4 - 4e^{-4} \right)$

Combining these results gives:

$V = \pi \left( \frac{1}{2} (e^8 - 1 + 32 - 8e^{-4}) \right)$

Thus, the exact value of the volume of the solid generated is:

$V = \frac{\pi}{2} (e^8 + 31 - 8e^{-4})$ and this is in its simplest form.