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Date	May 2019	Marks available	2	Reference code	19M.1.SL.TZ2.S_10
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Write down	Question number	S_10	Adapted from	N/A

Question

Let $y = {\left( {{x^3} + x} \right)^{\frac{3}{2}}}$ .

Consider the functions $f\left( x \right) = \sqrt {{x^3} + x}$ and $g\left( x \right) = 6 - 3{x^2}\sqrt {{x^3} + x}$ , for $x$ ≥ 0.

The graphs of $f$ and $g$ are shown in the following diagram.

The shaded region $R$ is enclosed by the graphs of $f$ , $g$ , the $y$ -axis and $x = 1$ .

Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .

[3]

Hence find $\int {\left( {3{x^2} + 1} \right)\sqrt {{x^3} + x} } \,{\text{d}}x$ .

[3]

Write down an expression for the area of $R$ .

[2]

Hence find the exact area of $R$ .

[6]

Markscheme

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

evidence of choosing chain rule (M1)

eg $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}}$ , $u = {x^3} + x$ , $u' = 3{x^2} + 1$

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{3}{2}{\left( {{x^3} + x} \right)^{\frac{1}{2}}}\left( {3{x^2} + 1} \right)\,\,\,\left( { = \frac{3}{2}\sqrt {{x^3} + x} \left( {3{x^2} + 1} \right)} \right)$ A2 N3

[3 marks]

integrating by inspection from (a) or by substitution (M1)

eg $\frac{2}{3}\int {\frac{3}{2}} \left( {3{x^2} + 1} \right)\sqrt {{x^3} + x} \,{\text{d}}x$ , $u = {x^3} + x$ , $\frac{{{\text{d}}u}}{{{\text{d}}x}} = 3{x^2} + 1$ , $\int {{u^{\frac{1}{2}}}}$ , $\frac{{{u^{\frac{3}{2}}}}}{{1.5}}$

correct integrated expression in terms of $x$ A2 N3

eg $\frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}} + C$ , $\frac{{{{\left( {{x^3} + x} \right)}^{1.5}}}}{{1.5}} + C$

[3 marks]

integrating and subtracting functions (in any order) (M1)

eg $\int {g - f}$ , $\int {f - \int g }$

correct integral (including limits, accept absence of ${{\text{d}}x}$ ) A1 N2

eg $\int_0^1 {\left( {g - f} \right)} \,{\text{d}}x$ , $\int_0^1 {6 - 3{x^2}\sqrt {{x^3} + x} - \sqrt {{x^3} + x} } \,{\text{d}}x$ , $\int_0^1 {g\left( x \right) - } \int_0^1 {f\left( x \right)}$

[2 marks]

recognizing $\sqrt {{x^3} + x}$ is a common factor (seen anywhere, may be seen in part (c)) (M1)

eg $\left( { - 3{x^2} - 1} \right)\sqrt {{x^3} + x}$ , $\int {6 - \left( {3{x^2} + 1} \right)} \sqrt {{x^3} + x}$ , $\left( {3{x^2} - 1} \right)\sqrt {{x^3} + x}$

correct integration (A1)(A1)

eg $6x - \frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}}$

Note: Award A1 for $6x$ and award A1 for $- \frac{2}{3}{\left( {{x^3} + x} \right)^{\frac{3}{2}}}$ .

substituting limits into their integrated function and subtracting (in any order) (M1)

eg $6 - \frac{2}{3}{\left( {{1^3} + 1} \right)^{\frac{3}{2}}}$ , $0 - \left[ {6 - \frac{2}{3}{{\left( {{1^3} + 1} \right)}^{\frac{3}{2}}}} \right]$

correct working (A1)

eg $6 - \frac{2}{3} \times 2\sqrt 2$ , $6 - \frac{2}{3} \times \sqrt 4 \times \sqrt 2$

area of $R = 6 - \frac{{4\sqrt 2 }}{3}\,\,\,\left( { = 6 - \frac{2}{3}\sqrt 8 {\text{,}}\,\,\,6 - \frac{2}{3} \times {2^{\frac{3}{2}}}{\text{,}}\,\,\,\frac{{18 - 4\sqrt 2 }}{3}} \right)$ A1 N3

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » AHL 5.11—Indefinite integration, reverse chain, by substitution

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