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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ2.6
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Hence and Determine	Question number	6	Adapted from	N/A

Question

The graphs of $y = 6 - x$ and $y = 1.5 x^{2} - 2.5 x + 3$ intersect at $(2, 4)$ and $(- 1, 7)$ , as shown in the following diagrams.

In diagram 1, the region enclosed by the lines $y = 6 - x$ , $x = - 1$ , $x = 2$ and the $x$ -axis has been shaded.

In diagram 2, the region enclosed by the curve $y = 1.5 x^{2} - 2.5 x + 3$ , and the lines $x = - 1$ , $x = 2$ and the $x$ -axis has been shaded.

Calculate the area of the shaded region in diagram 1.

[2]

Write down an integral for the area of the shaded region in diagram 2.

[2]

b.i.

Calculate the area of this region.

[1]

b.ii.

Hence, determine the area enclosed between $y = 6 - x$ and $y = 1.5 x^{2} - 2.5 x + 3$ .

[2]

Markscheme

EITHER

attempt to substitute $3, 4$ and $7$ into area of a trapezoid formula (M1)

$(A =) \frac{1}{2} (7 + 4) (3)$

given line expressed as an integral (M1)

$(A =) \int_{- 1}^{2} (6 - x) d x$

attempt to sum area of rectangle and area of triangle (M1)

$(A =) 4 \times 3 + \frac{1}{2} (3) (3)$

THEN

$16.5$ (square units) A1

[3 marks]

$(A =) \int_{- 1}^{2} 1.5 x^{2} - 2.5 x + 3 d x$ A1A1

Note: Award A1 for the limits $x = - 1$ , $x = 2$ in correct location. Award A1 for an integral of the quadratic function, $d x$ must be included. Do not accept “ $y$ ” in place of the function, given that two equations are in the question.

[2 marks]

b.i.

$9.75$ (square units) A1

[1 mark]

b.ii.

$16.5 - 9.75$ (M1)

$6.75$ (square units) A1

[2 marks]

Examiners report

There were a variety of methods used or attempted – area of trapezoid, integration, area of triangle plus area of rectangle, area of large rectangle minus area of top triangle, trapezoidal rule. All these methods, except for trapezoidal rule, proved successful for candidates, with the most common being the use of integration.

This was reasonably well done except for a few notation issues such as not including dx with their integrand. Those who attempted integration manually were not successful.

b.i.

[N/A]

b.ii.

Recognition that areas had to be subtracted was very evident.

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis

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