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Date	November 2018	Marks available	8	Reference code	18N.1.SL.TZ0.S_6
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	S_6	Adapted from	N/A

Question

Let $f\left( x \right) = \frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }}$ . The following diagram shows part of the graph of $f$ .

The region R is enclosed by the graph of $f$ , the $x$ -axis, and the $y$ -axis. Find the area of R.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (limits in terms of $x$ )

valid approach to find $x$ -intercept (M1)

eg $f\left( x \right) = 0$ , $\frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }} = 0$ , $6 - 2x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6x - {x^2}$ , $\int_0^3 {\frac{{6 - 2x}}{{\sqrt u }}} {\text{d}}x$ , ${\text{d}}u = 6 - 2x$ , $\int {\frac{1}{{\sqrt u }}}$ , $2{u^{\frac{1}{2}}}$ ,

$u = \sqrt {16 + 6x - {x^2}}$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = \left( {6 - 2x} \right)\frac{1}{2}{\left( {16 + 6x - {x^2}} \right)^{ - \frac{1}{2}}}$ , $\int 2 \,{\text{d}}u$ , $2u$

$\int {f\left( x \right)} \,{\text{d}}x = 2\sqrt {16 + 6x - {x^2}}$ (A2)

substituting both of their limits into their integrated function and subtracting (M1)

eg $2\sqrt {16 + 6\left( 3 \right) - {3^2}} - 2\sqrt {16 + 6{{\left( 0 \right)}^2} - {0^2}}$ , $2\sqrt {16 + 18 - 9} - 2\sqrt {16}$

Note: Award M0 if they substitute into original or differentiated function. Do not accept only “– 0” as evidence of substituting lower limit.

correct working (A1)

eg $2\sqrt {25} - 2\sqrt {16}$ , $10 - 8$

area = 2 A1 N2

METHOD 2 (limits in terms of $u$ )

valid approach to find $x$ -intercept (M1)

eg $f\left( x \right) = 0$ , $\frac{{6 - 2x}}{{\sqrt {16 + 6x - {x^2}} }} = 0$ , $6 - 2x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6x - {x^2}$ , $\int_0^3 {\frac{{6 - 2x}}{{\sqrt u }}} {\text{d}}x$ , ${\text{d}}u = 6 - 2x$ , $\int {\frac{1}{{\sqrt u }}}$ ,

$u = \sqrt {16 + 6x - {x^2}}$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = \left( {6 - 2x} \right)\frac{1}{2}{\left( {16 + 6x - {x^2}} \right)^{ - \frac{1}{2}}}$ , $\int 2 \,{\text{d}}u$

correct integration (A2)

eg $\int {\frac{1}{{\sqrt u }}} \,{\text{d}}u = 2{u^{\frac{1}{2}}}$ , $\int 2 \,{\text{d}}u = 2u$

both correct limits for $u$ (A1)

eg $u$ = 16 and $u$ = 25, $\int_{16}^{25} {\frac{1}{{\sqrt u }}{\text{d}}u}$ , $\left[ {2{u^{\frac{1}{2}}}} \right]_{16}^{25}$ , $u$ = 4 and $u$ = 5, $\int_4^5 2 \,{\text{d}}u$ , $\left[ {2u} \right]_4^5$

substituting both of their limits for $u$ (do not accept 0 and 3) into their integrated function and subtracting (M1)

eg $2\sqrt {25} - 2\sqrt {16}$ , $10 - 8$

Note: Award M0 if they substitute into original or differentiated function, or if they have not attempted to find limits for $u$ .

area = 2 A1 N2

[8 marks]

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Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

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