Figure 1 shows a sketch of part of the curve with equation $y = x^2 ext{ln} x$, $x > 1$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

The trapezium rule estimates the area under a curve by dividing it into trapezoids. The formula for the trapezium rule is:

$A = \frac{h}{2} \left( y_0 + 2 \sum_{i=1}^{n-1} y_i + y_n \right)$

where $h$ is the width of each segment and $y_i$ are the corresponding heights.

Here, we have six intervals from $x=1$ to $x=2$ with:

• $h=0.2$ (the width of each segment)
• values of $y$: $0$, $0.2625$, $0.6595$, $1.2032$, $1.9044$, $2.7726$.

Thus, applying the trapezium rule:

A & = \frac{0.2}{2} \left(0 + 2(0.2625 + 0.6595 + 1.2032 + 1.9044) + 2.7726 \right) \ & = 0.1 \left(0 + 2(4.0296) + 2.7726 \right) \ & = 0.1 \left(8.0592 + 2.7726 \right) \ & = 0.1 \times 10.8318 \ & = 1.0832.\end{align*}$$ Rounding to 3 decimal places, we find: $$A \approx 1.083.$$

To find the area under the curve using integration, we perform the definite integral of the function $y = x^2 ext{ln} x$ from $x = 1$ to $x = 2$.

The area can be calculated as:

$A = \int_{1}^{2} x^2 \ln x \, dx.$

This integral can be solved using integration by parts. Let:

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

Applying integration by parts:

$\int u \, dv = uv - \int v \, du$

Thus,

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

Calculate the first term:

\approx 1.5392.$$ Now calculate the integral: $$\int_{1}^{2} \frac{x^2}{3} \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{1}{3} \left(\frac{8}{3} - \frac{1}{3} \right) = \frac{1}{3} \times \frac{7}{3} = \frac{7}{9}.$$ Putting it all together: $$A = 1.5392 - \frac{7}{9} \approx 1.0805.$$ Thus, the exact value for the area $R$ is: $$A \approx 1.080.$$