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 6 - 2017 - Paper 5

First, expand the integrand:

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

So, the integral to evaluate is:

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

Next, evaluate each term in the integral separately:

1. The integral of $e^{2x}$: $\int e^{2x} \, dx = \frac{1}{2} e^{2x}$.

2. The integral of $4$: $\int 4 \, dx = 4x$.

3. The integral of $4e^{-x}$: $\int 4e^{-x} \, dx = -4e^{-x}$.

4. The integral of $4e^{-2x}$: $\int 4e^{-2x} \, dx = -2e^{-2x}$.

Now, combining the results, we have:

$V = \pi \left[ \frac{1}{2} e^{2x} + 4x - 4e^{-x} - 2e^{-2x} \right]_{0}^{4}$.

Plugging in the limits:

At $x = 4$:

• $e^{2 \times 4} = e^8$
• $4 \times 4 = 16$
• $4 e^{-4}$
• $2 e^{-8}$

At $x = 0$:

• $e^{0} = 1$
• $0$
• $-4$
• $-2$

So, $V = \pi \left[ \left( \frac{1}{2} e^8 + 16 - 4e^{-4} - 2e^{-8} \right) - \left( \frac{1}{2} + 0 + 4 + 2 \right) \right].$

After simplifying the expression, the final volume $V$ becomes:

$V = \pi \left[ \frac{1}{2} e^8 + 16 - 4e^{-4} - 2e^{-8} - \frac{1}{2} - 6 \right].$

Calculating this gives:

$V = \pi \left( \frac{1}{2} e^8 + 10 - 4e^{-4} - 2e^{-8} \right).$

Thus, the volume of the solid generated is: $V = \pi \left( \frac{1}{2} e^8 + 10 - 4e^{-4} - 2e^{-8} \right)$.