The diagram shows part of the graph of $y = e^{-x^2}$ - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 3

To estimate the area under the curve using the trapezium rule with 4 strips between $x = 0.1$ and $x = 0.5$, we calculate the width of each strip:

$h = \frac{0.5 - 0.1}{4} = 0.1$

Next, we find the values of the function at the boundaries and the intervals:

• $f(0.1) = e^{-0.1^2} \\approx 0.904837$
• $f(0.2) = e^{-0.2^2} \\approx 0.818731$
• $f(0.3) = e^{-0.3^2} \\approx 0.740818$
• $f(0.4) = e^{-0.4^2} \\approx 0.670320$
• $f(0.5) = e^{-0.5^2} \\approx 0.606531$

Using the trapezium rule formula:

$\text{Area} \approx \frac{h}{2} (f(a) + 2f_1 + 2f_2 + 2f_3 + f(b))$

Where:

• $f(a) = f(0.1)$
• $f(b) = f(0.5)$

The estimated area is:

$\approx \frac{0.1}{2} \left(0.904837 + 2(0.818731 + 0.740818 + 0.670320) + 0.606531 \right) \approx 0.361$

In part (a), we found that the function is concave on the interval $x \in \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. However, within the range from $x = 0.1$ to $x = 0.5$, the graph is entirely concave down. This means that the trapezium rule, which approximates the area under the curve using linear segments, will always under-estimate the true area beneath a concave curve, as it does not account for the curvature. Therefore, the estimate calculated in part (b) is an underestimate.

To find an upper estimate using a rectangle, we can consider the rectangle that has the same width as the interval, from $x = 0.1$ to $x = 0.5$, with the height equal to the maximum value of the function within this range, which occurs at $x = 0.1$:

$\text{Height} = f(0.1) = 0.904837$

Calculating the area of the rectangle:

$\text{Area} = \text{Width} \times \text{Height} = (0.5 - 0.1) \times 0.904837 = 0.4 \times 0.904837 \approx 0.36193748$

Combined with the previous estimate from part (b), we can see:

It shows the shaded area approximates to:

$0.36 < 0.4 < 0.40$

Thus, the shaded area is 0.4 correct to 1 decimal place.