The graph of $y = \frac{2x^3}{x^2 + 1}$ is shown for $0 \leq x \leq 4$ - AQA - A-Level Maths Mechanics - Question 14 - 2019 - Paper 1

To calculate the area under the curve using the trapezium rule with $n = 4$, we first need to determine the width of each strip:

$h = \frac{b - a}{n} = \frac{4 - 0}{4} = 1$

Next, we evaluate the function at the required points:

• $y(0) = \frac{2(0)^3}{(0)^2 + 1} = 0$
• $y(1) = \frac{2(1)^3}{(1)^2 + 1} = 1$
• $y(2) = \frac{2(2)^3}{(2)^2 + 1} = 3.2$
• $y(3) = \frac{2(3)^3}{(3)^2 + 1} = 5.4$
• $y(4) = \frac{2(4)^3}{(4)^2 + 1} = 8$

Now, apply the trapezium rule:

$A \approx \frac{h}{2} \left( y(0) + 2y(1) + 2y(2) + 2y(3) + y(4) \right)$
$A \approx \frac{1}{2} \left( 0 + 2(1) + 2(3.2) + 2(5.4) + 8 \right)$
$A \approx \frac{1}{2} \left( 0 + 2 + 6.4 + 10.8 + 8 \right) = \frac{1}{2} \left( 27.2 \right) = 13.6$

Thus, the area that Caroline should obtain is approximately $13.60$.

To find the exact area, we will calculate the integral of the function from $0$ to $4$:

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

Using the substitution $u = x^2 + 1$, we have $du = 2x \, dx$, which gives $dx = \frac{du}{2x}$. We also need to change the limits of integration:

When $x=0$, $u=1$; and when $x=4$, $u=17$.

Substituting, we find:

$A = \int_1^{17} \frac{2(x^2)(du)}{(u)}$
Since $x^2 = u-1$, we can rewrite the integral as:

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

Integrating gives us:

$u - \ln|u| \bigg|_1^{17} = \left( 17 - \ln(17) \right) - \left( 1 - \ln(1) \right) = 17 - \ln(17) - 1 = 16 - \ln(17)$

Thus, we have shown that the exact area A is $16 - \ln 17$.

As $n$ approaches infinity, the trapezium rule will yield a more accurate approximation of the area under the curve. Each trapezium will become narrower, meaning the sum of their areas will converge closer to the actual area. Therefore, as $n \to \infty$, Caroline's answer to part (a)(ii) will tend to the exact area, which we calculated to be $16 - \ln 17$.