IBDP MAI : Topic 2 Function - AHL 2.10 Scaling very large or small numbers using logarithms AI HL Paper 3
Question
This question examines methods to determine the area bounded by an unknown curve.
The curve \( y = f(x) \) is shown in the graph below, for \( 0 \leq x \leq 4.4 \).
The curve \( y = f(x) \) passes through the following points:
| \( x \) | 0 | 1.1 | 2.2 | 3.3 | 4.4 |
| \( y \) | 2 | 5 | 15 | 47 | 148 |
The task is to find the area bounded by the curve, the \( x \)-axis, the \( y \)-axis, and the line \( x = 4.4 \).
A possible model for the curve \( y = f(x) \) is a cubic function.
Another possible model for the curve \( y = f(x) \) is an exponential function, \( y = p e^{qx} \), where \( p, q \in \mathbb{R} \).
(a)
(i) Use the trapezoidal rule to find an estimate for the area. [3]
(ii) With reference to the shape of the graph, explain whether your answer to part (a)(i) is an overestimate or underestimate of the area. [2]
(i) Use the trapezoidal rule to find an estimate for the area. [3]
(ii) With reference to the shape of the graph, explain whether your answer to part (a)(i) is an overestimate or underestimate of the area. [2]
(b)
(i) Use all the coordinates in the table to find the equation of the least squares cubic regression curve. [3]
(ii) Write down the coefficient of determination. [1]
(i) Use all the coordinates in the table to find the equation of the least squares cubic regression curve. [3]
(ii) Write down the coefficient of determination. [1]
(c)
(i) Write down an expression for the area enclosed by the cubic function, the \( x \)-axis, the \( y \)-axis, and the line \( x = 4.4 \). [2]
(ii) Find the value of this area. [2]
(i) Write down an expression for the area enclosed by the cubic function, the \( x \)-axis, the \( y \)-axis, and the line \( x = 4.4 \). [2]
(ii) Find the value of this area. [2]
(d)
(i) Show that \( \ln y = q x + \ln p \). [2]
(ii) Hence explain how a straight line graph could be drawn using the coordinates in the table. [1]
(iii) By finding the equation of a suitable regression line, show that \( p = 1.83 \) and \( q = 0.986 \). [5]
(iv) Hence find the area enclosed by the exponential function, the \( x \)-axis, the \( y \)-axis, and the line \( x = 4.4 \). [2]
(i) Show that \( \ln y = q x + \ln p \). [2]
(ii) Hence explain how a straight line graph could be drawn using the coordinates in the table. [1]
(iii) By finding the equation of a suitable regression line, show that \( p = 1.83 \) and \( q = 0.986 \). [5]
(iv) Hence find the area enclosed by the exponential function, the \( x \)-axis, the \( y \)-axis, and the line \( x = 4.4 \). [2]
▶️ Answer/Explanation
Markscheme
(a)(i)
Using the trapezoidal rule with points \( (0, 2), (1.1, 5), (2.2, 15), (3.3, 47), (4.4, 148) \), width \( h = 1.1 \): M1
Area \( \approx \dfrac{1.1}{2} \left( 2 + 2(5 + 15 + 47) + 148 \right) \). A1
\( = \dfrac{1.1}{2} \times (2 + 134 + 148) = 0.55 \times 284 = 156 \) units\(^2\). A1
[3 marks]
Using the trapezoidal rule with points \( (0, 2), (1.1, 5), (2.2, 15), (3.3, 47), (4.4, 148) \), width \( h = 1.1 \): M1
Area \( \approx \dfrac{1.1}{2} \left( 2 + 2(5 + 15 + 47) + 148 \right) \). A1
\( = \dfrac{1.1}{2} \times (2 + 134 + 148) = 0.55 \times 284 = 156 \) units\(^2\). A1
[3 marks]
(a)(ii)
The graph is concave up. R1
Thus, the trapezoidal rule gives an overestimate of the area. A1
[2 marks]
The graph is concave up. R1
Thus, the trapezoidal rule gives an overestimate of the area. A1
[2 marks]
(b)(i)
Using least squares regression with points \( (0, 2), (1.1, 5), (2.2, 15), (3.3, 47), (4.4, 148) \): M1
Cubic function: \( f(x) = 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \). A2
[3 marks]
Using least squares regression with points \( (0, 2), (1.1, 5), (2.2, 15), (3.3, 47), (4.4, 148) \): M1
Cubic function: \( f(x) = 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \). A2
[3 marks]
(b)(ii)
Coefficient of determination: \( R^2 = 0.999 \). A1
[1 mark]
Coefficient of determination: \( R^2 = 0.999 \). A1
[1 mark]
(c)(i)
Area expression: \( \int_0^{4.4} \left( 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \right) dx \). A1 A1
[2 marks]
Area expression: \( \int_0^{4.4} \left( 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \right) dx \). A1 A1
[2 marks]
(c)(ii)
Integrate: \( \int \left( 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \right) dx = 0.97 x^4 – \dfrac{12.8}{3} x^3 + \dfrac{14.1}{2} x^2 + 1.54 x \).
Evaluate from 0 to 4.4: \( \left[ 0.97 (4.4)^4 – \dfrac{12.8}{3} (4.4)^3 + \dfrac{14.1}{2} (4.4)^2 + 1.54 \times 4.4 \right] – 0 \approx 145 \) units\(^2\). A2
[2 marks]
Integrate: \( \int \left( 3.88 x^3 – 12.8 x^2 + 14.1 x + 1.54 \right) dx = 0.97 x^4 – \dfrac{12.8}{3} x^3 + \dfrac{14.1}{2} x^2 + 1.54 x \).
Evaluate from 0 to 4.4: \( \left[ 0.97 (4.4)^4 – \dfrac{12.8}{3} (4.4)^3 + \dfrac{14.1}{2} (4.4)^2 + 1.54 \times 4.4 \right] – 0 \approx 145 \) units\(^2\). A2
[2 marks]
(d)(i)
For \( y = p e^{qx} \): M1
\( \ln y = \ln (p e^{qx}) = \ln p + \ln (e^{qx}) = \ln p + q x \).
Thus, \( \ln y = q x + \ln p \). A1
[2 marks]
For \( y = p e^{qx} \): M1
\( \ln y = \ln (p e^{qx}) = \ln p + \ln (e^{qx}) = \ln p + q x \).
Thus, \( \ln y = q x + \ln p \). A1
[2 marks]
(d)(ii)
Plot \( \ln y \) against \( x \), as \( \ln y = q x + \ln p \) represents a straight line with slope \( q \) and intercept \( \ln p \). R1
[1 mark]
Plot \( \ln y \) against \( x \), as \( \ln y = q x + \ln p \) represents a straight line with slope \( q \) and intercept \( \ln p \). R1
[1 mark]
(d)(iii)
Using points \( (x, \ln y) \): \( (0, \ln 2), (1.1, \ln 5), (2.2, \ln 15), (3.3, \ln 47), (4.4, \ln 148) \). M1
Linear regression: \( \ln y = 0.986 x + 0.602 \). A1
Slope: \( q = 0.986 \). R1
Intercept: \( \ln p = 0.602 \). M1
Thus, \( p = e^{0.602} \approx 1.83 \). A1
[5 marks]
Using points \( (x, \ln y) \): \( (0, \ln 2), (1.1, \ln 5), (2.2, \ln 15), (3.3, \ln 47), (4.4, \ln 148) \). M1
Linear regression: \( \ln y = 0.986 x + 0.602 \). A1
Slope: \( q = 0.986 \). R1
Intercept: \( \ln p = 0.602 \). M1
Thus, \( p = e^{0.602} \approx 1.83 \). A1
[5 marks]
(d)(iv)
For \( y = 1.83 e^{0.986 x} \), area = \( \int_0^{4.4} 1.83 e^{0.986 x} dx \). M1
\( \int e^{0.986 x} dx = \dfrac{1}{0.986} e^{0.986 x} \), so area = \( 1.83 \times \dfrac{1}{0.986} \left[ e^{0.986 \times 4.4} – e^0 \right] \approx 1.857 \times (e^{4.3384} – 1) \approx 141 \) units\(^2\). A1
[2 marks]
For \( y = 1.83 e^{0.986 x} \), area = \( \int_0^{4.4} 1.83 e^{0.986 x} dx \). M1
\( \int e^{0.986 x} dx = \dfrac{1}{0.986} e^{0.986 x} \), so area = \( 1.83 \times \dfrac{1}{0.986} \left[ e^{0.986 \times 4.4} – e^0 \right] \approx 1.857 \times (e^{4.3384} – 1) \approx 141 \) units\(^2\). A1
[2 marks]
Total Marks: 21