IB Mathematics SL 5.5 integration as anti-differentiation AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Olivia examines the curves of \( y = 6 – x \) and \( y = 1.5x^2 – 2.5x + 3 \), which intersect at (2, 4) and (-1, 7), as shown in the following diagrams. In diagram 1, the region bounded by the lines \( y = 6 – x \), \( x = -1 \), \( x = 2 \), and the x-axis is highlighted.
(a) Calculate the area of the highlighted region in diagram 1. [2]
In diagram 2, the region bounded by the curve \( y = 1.5x^2 – 2.5x + 3 \), the lines \( x = -1 \), \( x = 2 \), and the x-axis is highlighted.
(b)(i) Write down an integral for the area of the highlighted region in diagram 2. [2]
(b)(ii) Calculate the area of this region. [1]
(c) Hence, determine the area bounded between \( y = 6 – x \) and \( y = 1.5x^2 – 2.5x + 3 \). [2]
▶️ Answer/Explanation
Markscheme
(a)
The highlighted region is bounded by \( y = 6 – x \), \( x = -1 \), \( x = 2 \), and the x-axis. Area:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} (6 – x) \, dx \\ \int (6 – x) \, dx &= 6x – \frac{x^2}{2} + C \\ \left[ 6x – \frac{x^2}{2} \right]_{-1}^{2} &= \left( 6 \times 2 – \frac{2^2}{2} \right) – \left( 6 \times (-1) – \frac{(-1)^2}{2} \right) \\ &= (12 – 2) – \left( -6 – \frac{1}{2} \right) = 10 – (-6.5) = 10 + 6.5 = 16.5 \end{aligned} \]
Answer: \( 16.5 \, \text{square units} \) M1 A1
[2 marks]
The highlighted region is bounded by \( y = 6 – x \), \( x = -1 \), \( x = 2 \), and the x-axis. Area:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} (6 – x) \, dx \\ \int (6 – x) \, dx &= 6x – \frac{x^2}{2} + C \\ \left[ 6x – \frac{x^2}{2} \right]_{-1}^{2} &= \left( 6 \times 2 – \frac{2^2}{2} \right) – \left( 6 \times (-1) – \frac{(-1)^2}{2} \right) \\ &= (12 – 2) – \left( -6 – \frac{1}{2} \right) = 10 – (-6.5) = 10 + 6.5 = 16.5 \end{aligned} \]
Answer: \( 16.5 \, \text{square units} \) M1 A1
[2 marks]
(b)(i)
The highlighted region is bounded by \( y = 1.5x^2 – 2.5x + 3 \), \( x = -1 \), \( x = 2 \), and the x-axis. Area:
\[ \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \]
Answer: \( \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \) A1 A1
[2 marks]
The highlighted region is bounded by \( y = 1.5x^2 – 2.5x + 3 \), \( x = -1 \), \( x = 2 \), and the x-axis. Area:
\[ \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \]
Answer: \( \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \) A1 A1
[2 marks]
(b)(ii)
Compute:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \\ \int (1.5x^2 – 2.5x + 3) \, dx &= 0.5x^3 – 1.25x^2 + 3x + C \\ \left[ 0.5x^3 – 1.25x^2 + 3x \right]_{-1}^{2} &= \left( 0.5 \times 2^3 – 1.25 \times 2^2 + 3 \times 2 \right) – \left( 0.5 \times (-1)^3 – 1.25 \times (-1)^2 + 3 \times (-1) \right) \\ &= (4 – 5 + 6) – (-0.5 – 1.25 – 3) = 5 – (-4.75) = 5 + 4.75 = 9.75 \end{aligned} \]
Answer: \( 9.75 \, \text{square units} \) A1
[1 mark]
Compute:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \\ \int (1.5x^2 – 2.5x + 3) \, dx &= 0.5x^3 – 1.25x^2 + 3x + C \\ \left[ 0.5x^3 – 1.25x^2 + 3x \right]_{-1}^{2} &= \left( 0.5 \times 2^3 – 1.25 \times 2^2 + 3 \times 2 \right) – \left( 0.5 \times (-1)^3 – 1.25 \times (-1)^2 + 3 \times (-1) \right) \\ &= (4 – 5 + 6) – (-0.5 – 1.25 – 3) = 5 – (-4.75) = 5 + 4.75 = 9.75 \end{aligned} \]
Answer: \( 9.75 \, \text{square units} \) A1
[1 mark]
(c)
The area between \( y = 6 – x \) and \( y = 1.5x^2 – 2.5x + 3 \) is the area under the line minus the area under the curve:
\[ \text{Area} = 16.5 – 9.75 = 6.75 \]
Alternative method:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} \left[ (6 – x) – (1.5x^2 – 2.5x + 3) \right] \, dx = \int_{-1}^{2} (-1.5x^2 + 1.5x + 3) \, dx \\ \int (-1.5x^2 + 1.5x + 3) \, dx &= -0.5x^3 + 0.75x^2 + 3x + C \\ \left[ -0.5x^3 + 0.75x^2 + 3x \right]_{-1}^{2} &= \left( -0.5 \times 2^3 + 0.75 \times 2^2 + 3 \times 2 \right) – \left( -0.5 \times (-1)^3 + 0.75 \times (-1)^2 + 3 \times (-1) \right) \\ &= (-4 + 3 + 6) – (0.5 + 0.75 – 3) = 5 – (-1.75) = 5 + 1.75 = 6.75 \end{aligned} \]
Answer: \( 6.75 \, \text{square units} \) M1 A1
[2 marks]
The area between \( y = 6 – x \) and \( y = 1.5x^2 – 2.5x + 3 \) is the area under the line minus the area under the curve:
\[ \text{Area} = 16.5 – 9.75 = 6.75 \]
Alternative method:
\[ \begin{aligned} \text{Area} &= \int_{-1}^{2} \left[ (6 – x) – (1.5x^2 – 2.5x + 3) \right] \, dx = \int_{-1}^{2} (-1.5x^2 + 1.5x + 3) \, dx \\ \int (-1.5x^2 + 1.5x + 3) \, dx &= -0.5x^3 + 0.75x^2 + 3x + C \\ \left[ -0.5x^3 + 0.75x^2 + 3x \right]_{-1}^{2} &= \left( -0.5 \times 2^3 + 0.75 \times 2^2 + 3 \times 2 \right) – \left( -0.5 \times (-1)^3 + 0.75 \times (-1)^2 + 3 \times (-1) \right) \\ &= (-4 + 3 + 6) – (0.5 + 0.75 – 3) = 5 – (-1.75) = 5 + 1.75 = 6.75 \end{aligned} \]
Answer: \( 6.75 \, \text{square units} \) M1 A1
[2 marks]
Total Marks: 7
Question
Taylor hits golf balls into the air. Each time they hit a ball, they record \( \theta \), the angle at which the ball is launched into the air, and \( l \), the horizontal distance, in metres, which the ball travels from the point of contact to the first time it lands. The diagram below represents this information.
Taylor analyses their results and concludes: \( \dfrac{dl}{d\theta} = -0.2\theta + 9 \), \( 35^\circ \leq \theta \leq 75^\circ \).
(a) Determine whether the function \( l \) with respect to \( \theta \) is increasing or decreasing at \( \theta = 50^\circ \). [2]
(b) Taylor observes that when the angle is \( 40^\circ \), the ball travels a horizontal distance of 205.5 m. [2]
(c) Determine an expression for the function \( l(\theta) \). [4]
▶️ Answer/Explanation
Markscheme
(a)
Given: \( \dfrac{dl}{d\theta} = -0.2\theta + 9 \). M1
Evaluate at \( \theta = 50^\circ \): \( \dfrac{dl}{d\theta} = -0.2 \times 50 + 9 = -10 + 9 = -1 \).
Since \( \dfrac{dl}{d\theta} = -1 < 0 \), the function \( l \) is decreasing at \( \theta = 50^\circ \). A1
[2 marks]
Given: \( \dfrac{dl}{d\theta} = -0.2\theta + 9 \). M1
Evaluate at \( \theta = 50^\circ \): \( \dfrac{dl}{d\theta} = -0.2 \times 50 + 9 = -10 + 9 = -1 \).
Since \( \dfrac{dl}{d\theta} = -1 < 0 \), the function \( l \) is decreasing at \( \theta = 50^\circ \). A1
[2 marks]
(b)
Given: \( l(40) = 205.5 \, \text{m} \). A1
This information is used in part (c). A1
[2 marks]
Given: \( l(40) = 205.5 \, \text{m} \). A1
This information is used in part (c). A1
[2 marks]
(c)
Given: \( \dfrac{dl}{d\theta} = -0.2\theta + 9 \). M1
Integrate: \( l(\theta) = \int (-0.2\theta + 9) \, d\theta \).
Antiderivative: \( \int -0.2\theta \, d\theta = -0.2 \times \dfrac{\theta^2}{2} = -0.1\theta^2 \), \( \int 9 \, d\theta = 9\theta \).
General form: \( l(\theta) = -0.1\theta^2 + 9\theta + c \). A1
Use condition: \( l(40) = 205.5 \). Substitute: \( 205.5 = -0.1 \times 40^2 + 9 \times 40 + c \). M1
Calculate: \( 40^2 = 1600 \), \( -0.1 \times 1600 = -160 \), \( 9 \times 40 = 360 \), so \( 205.5 = -160 + 360 + c = 200 + c \).
Solve: \( c = 205.5 – 200 = 5.5 \).
Result: \( l(\theta) = -0.1\theta^2 + 9\theta + 5.5 \). A1
[4 marks]
Given: \( \dfrac{dl}{d\theta} = -0.2\theta + 9 \). M1
Integrate: \( l(\theta) = \int (-0.2\theta + 9) \, d\theta \).
Antiderivative: \( \int -0.2\theta \, d\theta = -0.2 \times \dfrac{\theta^2}{2} = -0.1\theta^2 \), \( \int 9 \, d\theta = 9\theta \).
General form: \( l(\theta) = -0.1\theta^2 + 9\theta + c \). A1
Use condition: \( l(40) = 205.5 \). Substitute: \( 205.5 = -0.1 \times 40^2 + 9 \times 40 + c \). M1
Calculate: \( 40^2 = 1600 \), \( -0.1 \times 1600 = -160 \), \( 9 \times 40 = 360 \), so \( 205.5 = -160 + 360 + c = 200 + c \).
Solve: \( c = 205.5 – 200 = 5.5 \).
Result: \( l(\theta) = -0.1\theta^2 + 9\theta + 5.5 \). A1
[4 marks]
Total Marks: 8