The curve shown in Figure 2 has equation $y = \frac{1}{2x+1}$ - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 8

To find the volume of the solid generated by rotating the region about the x-axis, we use the disk method. The volume $V$ can be expressed as:

$V = \pi \int_a^b \left( \frac{1}{2x + 1} \right)^2 dx$

Step 1: Set up the integral

We first set up the integral with respect to the variable x:

$V = \pi \int_a^b \frac{1}{(2x + 1)^2} dx$

Step 2: Evaluate the integral

To evaluate the integral, we can use the substitution method. Letting $u = 2x + 1$, then $du = 2dx$ or $dx = \frac{1}{2}du$. We need to change the limits accordingly:

• When $x = a$, $u = 2a + 1$.
• When $x = b$, $u = 2b + 1$.

Then, the integral becomes:

$V = \pi \int_{2a + 1}^{2b + 1} \frac{1}{u^2} \cdot \frac{1}{2} du$

= $\frac{\pi}{2} \int_{2a + 1}^{2b + 1} u^{-2} du$

Step 3: Integrate

Now we can integrate:

$V = \frac{\pi}{2}\left[ -\frac{1}{u} \right]_{2a + 1}^{2b + 1}$

This evaluates to:

$V = \frac{\pi}{2}\left( -\frac{1}{2b + 1} + \frac{1}{2a + 1} \right)$

Step 4: Calculate the final volume

Rearranging gives:

$V = \frac{\pi}{2} \left( \frac{1}{2a + 1} - \frac{1}{2b + 1} \right)$

Combining into a single fraction yields:

$V = \frac{\pi(2b + 1 - 2a - 1)}{2(2a + 1)(2b + 1)} = \frac{\pi(2b - 2a)}{2(2a + 1)(2b + 1)}$

Thus, the final answer can be expressed as:

$V = \frac{\pi(b - a)}{(2a + 1)(2b + 1)}$

This represents the volume of the solid generated, expressed as a single simplified fraction in terms of $a$ and $b$.