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Date May Example question Marks available 3 Reference code EXM.3.AHL.TZ0.7
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 7 Adapted from N/A

Question

This question explores methods to determine the area bounded by an unknown curve.

The curve  y = f ( x )  is shown in the graph, for  0 x 4.4 .

The curve y = f ( x )  passes through the following points.

It is required to find the area bounded by the curve, the  x -axis, the y -axis and the line  x = 4.4 .

One possible model for the curve  y = f ( x )  is a cubic function.

A second possible model for the curve  y = f ( x )  is an exponential function,  y = p e q x , where  p , q R .

Use the trapezoidal rule to find an estimate for the area.

[3]
a.i.

With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.

[2]
a.ii.

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

[3]
b.i.

Write down the coefficient of determination.

[1]
b.ii.

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]
c.i.

Find the value of this area.

[2]
c.ii.

Show that  ln y = q x + ln p .

[2]
d.i.

Hence explain how a straight line graph could be drawn using the coordinates in the table.

[1]
d.ii.

By finding the equation of a suitable regression line, show that  p = 1.83 and  q = 0.986 .

[5]
d.iii.

Hence find the area enclosed by the exponential function, the x -axis, the y -axis and the line x = 4.4 .

[2]
d.iv.

Markscheme

Area  = 1.1 2 ( 2 + 2 ( 5 + 15 + 47 ) + 148 )          M1A1

Area = 156 units2          A1

[3 marks]

a.i.

The graph is concave up,         R1

so the trapezoidal rule will give an overestimate.         A1

[2 marks]

a.ii.

f ( x ) = 3.88 x 3 12.8 x 2 + 14.1 x + 1.54          M1A2

[3 marks]

b.i.

R 2 = 0.999         A1

[1 mark]

b.ii.

Area  = 0 4.4 ( 3.88 x 3 12.8 x 2 + 14.1 x + 1.54 ) d x         A1A1

[2 marks]

c.i.

Area = 145 units2    (Condone 143–145 units2, using rounded values.)      A2

[2 marks]

c.ii.

ln y = ln ( p e q x )       M1

ln y = ln p + ln ( e q x )       A1

ln y = q x + ln p       AG

[2 marks]

d.i.

Plot  ln y against p .      R1

[1 mark]

d.ii.

Regression line is  ln y = 0.986 x + 0.602        M1A1

So  q = gradient = 0.986    R1

p = e 0.602 = 1.83        M1A1

[5 marks]

d.iii.

Area  = 0 4.4 1.83 e 0.986 x d x = 140  units2     M1A1

[2 marks]

d.iv.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
d.iii.
[N/A]
d.iv.

Syllabus sections

Topic 5—Calculus » SL 5.8—Trapezoid rule
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Topic 2—Functions » AHL 2.10—Scaling large numbers, log-log graphs
Topic 4—Statistics and probability » AHL 4.13—Non-linear regression
Topic 2—Functions
Topic 4—Statistics and probability
Topic 5—Calculus

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