Figure 2 shows a sketch of the curve with equation $y = x^3 \ln(x^2 + 2)$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 5

To estimate the area using the trapezium rule, we apply the formula:

$\text{Area} \approx \frac{1}{2} h (y_0 + 2y_1 + 2y_2 + y_3)$

where $h$ is the width of each interval and $y_i$ are the corresponding $y$ values. Here:

• $h = \frac{\sqrt{2}}{4} - 0 = \frac{\sqrt{2}}{4}$
• Values from the table are $y_0 = 0$, $y_1 = 0.3240$, $y_2 = 1.3596$, and $y_3 = 3.9210$.

Thus:

$\text{Area} \approx \frac{1}{2} \times \frac{\sqrt{2}}{4} \times (0 + 2 \times 0.3240 + 2 \times 1.3596 + 3.9210)$

Calculating this gives:

$\text{Area} \approx 1.30$

Using the substitution $u = x^2 + 2$, we find:

$\frac{du}{dx} = 2x \implies du = 2x \, dx \implies dx = \frac{du}{2x}$

Now we need to change the limits:

• When $x = 0$, $u = 2$.
• When $x = \sqrt{2}$, $u = 4$.

Thus, the area $A(R)$ can be expressed as:

$A(R) = \int_{0}^{\sqrt{2}} y \, dx = \int_{0}^{\sqrt{2}} x^3 \ln(x^2 + 2) \, dx$

Substituting $u$ into the integral gives:

$A(R) = \int_{2}^{4} \left(\frac{u-2}{2} \ln u \right) \frac{du}{2} = \frac{1}{2} \int_{2}^{4} (u - 2) \ln u \, du$

To find the exact area of $R$, we evaluate the integral:

$A(R) = \frac{1}{2} \int_{2}^{4} (u - 2) \ln u \, du$.

To solve this, we break it down:

1. Integrate using integration by parts, letting:
   • $v = \ln u$ and $dw = (u - 2) du$.

The solution yields:

$A(R) = \frac{1}{2} \Bigg[ (u - 2) \ln u - \int (1) \ln u \, du \Bigg]_{2}^{4}$

By evaluating this, we can find the exact area of $R$.