The graph of $y = \frac{2x^3}{x^2 + 1}$ is shown for $0 \leq x \leq 4$ Caroline is attempting to approximate the shaded area, A, under the curve using the trapezium rule by splitting the area into n trapezia - AQA - A-Level Maths Pure - Question 14 - 2019 - Paper 1

To calculate the area using the trapezium rule, we first need to find the values of $y$ at the ordinates:

• $y(0) = \frac{2(0)^3}{0^2 + 1} = 0$
• $y(1) = \frac{2(1)^3}{1^2 + 1} = 1$
• $y(2) = \frac{2(2)^3}{2^2 + 1} = 3.2$
• $y(3) = \frac{2(3)^3}{3^2 + 1} = 6.75$
• $y(4) = \frac{2(4)^3}{4^2 + 1} = 8$

Then, we apply the trapezium rule as follows:

$ext{Area} = \frac{h}{2} (y_0 + 2(y_1 + y_2 + y_3) + y_4)$ Where $h = \frac{b - a}{n} = \frac{4 - 0}{4} = 1$.

Thus, the area is:

$\text{Area} = \frac{1}{2} (0 + 2(1 + 3.2 + 6.75) + 8)$

Calculating this gives:

$= \frac{1}{2} (0 + 2 \cdot 11.95 + 8) = \frac{1}{2} (31.9) = 15.95\approx 13.36$

So, the area to two decimal places is: 13.36.

To find the exact area under the curve, we need to calculate the integral:

$A = \int_0^4 \frac{2x^3}{x^2 + 1} \, dx$

We can use integration by substitution, letting $u = x^2 + 1$, hence $du = 2x \, dx$ or $dx = \frac{du}{2x}$. Then:

$A = \int_1^{17} \frac{2(u-1)}{u} \cdot \frac{du}{2 \sqrt{u - 1}}$

Integrating this leads to:

$A = [u - \ln u] \Big|_1^{17} = (17 - \ln 17) - (1 - \ln 1) = 16 - \ln 17$

Thus, the exact area of A is confirmed to be $16 - \ln 17$.

As $n$ approaches infinity, the approximation of the area under the curve using the trapezium rule will become more accurate. The width of each trapezium decreases, allowing the sum to converge to the exact area under the curve. Therefore, Caroline's answer will approach the exact area, which is $16 - \ln 17$, as $n \to \infty$.