Figure 1 shows a sketch of part of the curve with equation $y = x^2 ext{ln} x$, \quad x > 1 The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line $x = 2$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

Using the trapezium rule, we calculate:

1. The width of each section (interval) is $h = 0.2$ (since $1.0$, $1.2$, $1.4$, $1.6$, $1.8$, $2.0$ gives 5 intervals of 0.2).
2. The trapezium rule formula is given by:

$ext{Area} \approx \frac{h}{2} \left( y_0 + 2(y_1 + y_2 + y_3 + y_4) + y_5 \right)$

Substituting the values into the formula:

$\text{Area} \approx \frac{0.2}{2} \left( 0 + 2(0.2625 + 0.6595 + 1.2032 + 1.9044) + 2.7726 \right)$

3. Simplifying

• First calculate the sum inside: $0.2625 + 0.6595 + 1.2032 + 1.9044 \approx 4.0296$.
• Now substituting that: $\text{Area} \approx 0.1 \left( 0 + 2(4.0296) + 2.7726 \right)$
• Which gives: $\text{Area} \approx 0.1(8.0592 + 2.7726) \approx 0.1(10.8318)\approx 1.0832$

Therefore, the estimated area of $R$ is approximately $1.083$.

To find the exact area of region $R$ using integration, we set up the integral:

$A = \int_1^2 x^2 \ln x \, dx$

Using integration by parts, let:

• $u = \ln x$ then $du = \frac{1}{x} \, dx$,
• and $dv = x^2 \, dx$ then $v = \frac{x^3}{3}$.

Applying integration by parts:

$A = \left[ \frac{x^3}{3} \ln x \right]_1^2 - \int_1^2 \frac{x^3}{3} \cdot \frac{1}{x} \, dx$

• Calculating the boundary term: $A = \left[ \frac{8}{3} \ln 2 - \frac{1}{3}(0) \right] - \int_1^2 \frac{x^2}{3} \, dx$
• Now calculate the integral: $\int_1^2 \frac{x^2}{3} \, dx = \left[ \frac{x^3}{9} \right]_1^2 = \frac{8}{9} - \frac{1}{9} = \frac{7}{9}$

Putting this all together: $A = \frac{8}{3} \ln 2 - \frac{7}{27}$

Hence, the exact area of $R$ is: $A \approx 1.083$ when evaluated numerically.