Question
The following diagram shows a model of the side view of a water slide. All lengths are measured in metres.
(a) Use the trapezoidal rule with an interval width of 1 to calculate the approximate area under the model of the slide in the interval \( 0 \leq x \leq 4 \).
(b) Find \( \int \left( -\frac{1}{4}x^{2} + 2x \right) dx \).
(c) Calculate the exact area under the entire model of the slide, for \( 0 \leq x \leq 12 \).
(d) Find the percentage error in the total area under the entire model of the slide when using the approximate value from part (a).
▶️Answer/Explanation
Detailed solution
(a) Approximate Area Using Trapezoidal Rule:
The trapezoidal rule approximates the area under a curve by dividing it into trapezoids. For the interval \( 0 \leq x \leq 4 \) with an interval width of 1, the \( x \)-values are \( 0, 1, 2, 3, 4 \). The corresponding heights (values of the function) are:
\[ f(0) = 0, \quad f(1) = 1.75, \quad f(2) = 3, \quad f(3) = 3.75, \quad f(4) = 4 \]
The trapezoidal rule formula is:
\[ \text{Area} = \frac{1}{2} \times \text{interval width} \times \left( \text{first height} + \text{last height} + 2 \times \text{sum of middle heights} \right) \]
Substitute the values:
\[ \text{Area} = \frac{1}{2} \times 1 \times \left( 0 + 4 + 2(1.75 + 3 + 3.75) \right) \]
\[ \text{Area} = \frac{1}{2} \times 1 \times \left( 4 + 2(8.5) \right) = \frac{1}{2} \times 1 \times (4 + 17) = \frac{1}{2} \times 21 = 10.5 \, \text{m}^2 \]
So, the approximate area is 10.5 m².
(b) Integral of \( -\frac{1}{4}x^{2} + 2x \):
To find the indefinite integral:
\[ \int \left( -\frac{1}{4}x^{2} + 2x \right) dx = -\frac{1}{4} \cdot \frac{x^{3}}{3} + 2 \cdot \frac{x^{2}}{2} + c = -\frac{1}{12}x^{3} + x^{2} + c \]
So, the integral is:
\[ -\frac{1}{12}x^{3} + x^{2} + c \]
(c) Exact Area Under the Entire Model (\( 0 \leq x \leq 12 \)):
The area under the curve is divided into three parts:
- From \( x = 0 \) to \( x = 4 \): The area is given by the integral of \( -\frac{1}{4}x^{2} + 2x \).
- From \( x = 4 \) to \( x = 8 \): The area is a rectangle with height 4 and width 4.
- From \( x = 8 \) to \( x = 12 \): The area is a triangle with base 4 and height 7.
1. Area from \( x = 0 \) to \( x = 4 \):
Evaluate the definite integral:
\[ \int_{0}^{4} \left( -\frac{1}{4}x^{2} + 2x \right) dx = \left[ -\frac{1}{12}x^{3} + x^{2} \right]_{0}^{4} \]
\[ = \left( -\frac{1}{12}(64) + 16 \right) – \left( 0 + 0 \right) = -\frac{64}{12} + 16 = -\frac{16}{3} + 16 = \frac{32}{3} \approx 10.6667 \, \text{m}^2 \]
2. Area from \( x = 4 \) to \( x = 8 \):
This is a rectangle with height 4 and width 4:
\[ \text{Area} = 4 \times 4 = 16 \, \text{m}^2 \]
3. Area from \( x = 8 \) to \( x = 12 \):
This is a triangle with base 4 and height 7:
\[ \text{Area} = \frac{1}{2} \times 4 \times 7 = 14 \, \text{m}^2 \]
Total Exact Area:
\[ \text{Total Area} = 10.6667 + 16 + 14 = 40.6667 \, \text{m}^2 \]
So, the exact area under the entire model is \( \frac{122}{3} \, \text{m}^2 \) or approximately 40.6667 m².
(d) Percentage Error:
The approximate total area using part (a) is:
\[ \text{Approximate Area} = 10.5 + 16 + 14 = 40.5 \, \text{m}^2 \]
The exact total area is \( \frac{122}{3} \, \text{m}^2 \approx 40.6667 \, \text{m}^2 \).
The percentage error is:
\[ \text{Percentage Error} = \left| \frac{\text{Approximate Area} – \text{Exact Area}}{\text{Exact Area}} \right| \times 100 \]
\[ \text{Percentage Error} = \left| \frac{40.5 – 40.6667}{40.6667} \right| \times 100 \approx 0.4098\% \]
So, the percentage error is approximately 0.41%.
……………………………Markscheme……………………………….
(a) Approximate area = \( 10.5 \, \text{m}^2 \).
(b) Integral = \( -\frac{1}{12}x^{3} + x^{2} + c \).
(c) Exact area = \( \frac{122}{3} \, \text{m}^2 \) or \( 40.\overline{6} \, \text{m}^2 \).
(d) Percentage error = \( 0.41\% \).