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 find the height at each ordinate, which occurs at $x = 0$, $1$, $2$, $3$, and $4$.

We calculate:

• For $x = 0$, $y = \frac{2(0)^3}{(0)^2 + 1} = 0$
• For $x = 1$, $y = \frac{2(1)^3}{(1)^2 + 1} = 1$
• For $x = 2$, $y = \frac{2(2)^3}{(2)^2 + 1} = 3.2$
• For $x = 3$, $y = \frac{2(3)^3}{(3)^2 + 1} = 5.4$
• For $x = 4$, $y = \frac{2(4)^3}{(4)^2 + 1} = 7.2$

Using the trapezium rule:

$ext{Area} = \frac{1}{2} h \left( y_0 + 2y_1 + 2y_2 + 2y_3 + y_4 \right)$

where $h = \frac{4-0}{4} = 1$. Thus,

$\text{Area} = \frac{1}{2} \times 1 \left( 0 + 2(1) + 2(3.2) + 2(5.4) + 7.2 \right) = \frac{1}{2} \times 1 \left( 0 + 2 + 6.4 + 10.8 + 7.2 \right)$

Calculating this:

$= \frac{1}{2} \times 26.4 = 13.2$

Thus, the area that Caroline should obtain is $13.2$.

To find the exact area A, we integrate the function:

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

Using the substitution $u = x^2 + 1$, we have $du = 2x \, dx$, thus:

• When $x = 0$, $u = 1$.
• When $x = 4$, $u = 17$.

Rearranging for $dx$ gives:

$dx = \frac{du}{2x} = \frac{du}{2 \sqrt{u-1}}$

Then the integral becomes:

$\int_1^{17} \frac{u-1}{u} \, du$

Calculating this:

$= \int_1^{17} \left(1 - \frac{1}{u}\right) du = \left[ u - \ln |u| \right]_1^{17}$

Evaluating this from 1 to 17 results in:

$= \left( 17 - \ln(17) \right) - \left( 1 - \ln(1) \right) = 16 - \ln(17)$

Thus, confirming that the exact area A is $16 - \ln(17)$.

As $n \to \infty$, the trapezium rule becomes a more accurate approximation for the definite integral. Thus, Caroline's estimate for the area will converge towards the actual area under the curve, which we determined to be $16 - \ln 17$. Therefore, the error in her approximation will tend to $0$ as she increases the number of trapezia.