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

The trapezium rule formula is given by:

[\int_{a}^{b} f(x) , dx \approx \frac{(b-a)}{n} \left( \frac{f(a) + f(b)}{2} + \sum_{i=1}^{n-1} f(x_i) \right)]

In this case, $a = 0.1$, $b = 0.5$, and $n = 4$.

The width of each strip is:

[h = \frac{0.5 - 0.1}{4} = 0.1]

Now we calculate the $x$ values:

• $x_0 = 0.1$, $f(0.1) = e^{-0.01} \approx 0.9900$
• $x_1 = 0.2$, $f(0.2) = e^{-0.04} \approx 0.9608$
• $x_2 = 0.3$, $f(0.3) = e^{-0.09} \approx 0.9139$
• $x_3 = 0.4$, $f(0.4) = e^{-0.16} \approx 0.8521$
• $x_4 = 0.5$, $f(0.5) = e^{-0.25} \approx 0.7788$

Applying the trapezium rule:

[\int_{0.1}^{0.5} e^{-x^2} , dx \approx \frac{0.4}{8} \left( 0.9900 + 0.7788 + 2(0.9608 + 0.9139 + 0.8521) \right)]

Calculating:

[\approx 0.1 \left( 0.9900 + 0.7788 + 2(0.9608 + 0.9139 + 0.8521) \right) \approx 0.3611]

Thus, the estimate to four decimal places is:

[\approx 0.3611]

Referring back to part (a), we established that the graph is concave between [-\frac{\sqrt{2}}{2}] and [\frac{\sqrt{2}}{2}]. Since the function is concave in the interval from $0.1$ to $0.5$, the trapezium rule will give an underestimate because the trapezoids constructed will lie entirely below the curve in a concave section.

To calculate the area of a rectangle that encloses the shaded region, we consider the width $w$ of the rectangle as [0.4] (since it runs from $0.1$ to $0.5$) and the height $h$ as given by the value of the function at the lower limit [f(0.1) \approx 0.9900]. Therefore, the area of the rectangle is:

[\text{Area} = w \times h = 0.4 \times 0.9900 = 0.396]

Now rounding to 1 decimal place gives us:[0.4]. Thus, we find that the shaded area is indeed 0.4 correct to 1 decimal place.