IB DP Maths Topic 6.5 Volumes of revolution about the x-axis SL Paper 1

Question

Let  \(f:x \mapsto {\sin ^3}x\) .

(i) Write down the range of the function f .

(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.

[5]
a.

Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .

[2]
b.

Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.

[7]
c.
Answer/Explanation

Markscheme

(i) range of f is \([ – 1{\text{, }}1]\) , \(( – 1 \le f(x) \le 1)\)     A2     N2

(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\)     A1

justification for one solution on \([0{\text{, }}2\pi ]\)    R1

e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)

1 solution (seen anywhere)     A1     N1

[5 marks]

a.

\(f'(x) = 3{\sin ^2}x\cos x\)     A2     N2

[2 marks]

b.

using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \)     (M1)

\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\)     (A1)

\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \)     A1

\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) – {{\sin }^3}0} \right)} \right)\)     A2

evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\)     (A1)

e.g. \(\pi \left( {1 – 0} \right)\)

\(V = \pi \)     A1     N1

[7 marks]

c.

Question

The graph of \(y = \sqrt x \) between \(x = 0\) and \(x = a\) is rotated \(360^\circ \) about the x-axis. The volume of the solid formed is \(32\pi \) . Find the value of a.

Answer/Explanation

Markscheme

attempt to substitute into formula \(V = \int {\pi {y^2}{\rm{d}}x} \)     (M1) 

integral expression     A1

e.g. \(\pi \int_0^a {(\sqrt x } {)^2}{\rm{d}}x\) ,  \(\pi \int x \)

correct integration     (A1)

e.g. \(\int x{{\rm{d}}x = \frac{1}{2}{x^2}}\)

correct substitution \(V = \pi \left[ {\frac{1}{2}{a^2}} \right]\)     (A1)

equating their expression to \(32\pi \)     M1

e.g. \(\pi \left[ {\frac{1}{2}{a^2}} \right] = 32\pi \)

\({a^2} = 64\)

\(a = 8\)     A2     N2

[7 marks]

Question

Let \(f(x) = \sqrt x \) . Line L is the normal to the graph of f at the point (4, 2) .

In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .


Show that the equation of L is \(y = – 4x + 18\) .

[4]
a.

Point A is the x-intercept of L . Find the x-coordinate of A.

[2]
b.

Find an expression for the area of R .

[3]
c.

The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .

[8]
d.
Answer/Explanation

Markscheme

finding derivative     (A1)

e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)

gradient of normal =  \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere)     A1

e.g. \( – \frac{1}{{f'(4)}} = – 4\) , \( – 2\sqrt x \)

substituting into equation of line (for normal)     M1

e.g. \(y – 2 = – 4(x – 4)\)

\(y = – 4x + 18\)     AG     N0

[4 marks]

a.

recognition that \(y = 0\) at A     (M1)

e.g. \( – 4x + 18 = 0\)

\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\)     A1     N2

[2 marks]

b.

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)

Note: Award A1 if dx is missing.

[3 marks]

c.

correct expression for the volume from \(x = 0\) to \(x = 4\)     (A1)

e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)

\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\)     A1

\(V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)\)     (A1)

\(V = 8\pi \)     A1

finding the volume from \(x = 4\) to \(x = 4.5\)

EITHER

recognizing a cone     (M1)

e.g. \(V = \frac{1}{3}\pi {r^2}h\)

\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\)     (A1)

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

OR

\(V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x} \)     (M1)

\( = \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x} \)

\( = \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}\)     A1

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

[8 marks]

d.

Question

The graph of \(f(x) = \sqrt {16 – 4{x^2}} \) , for \( – 2 \le x \le 2\) , is shown below.


The region enclosed by the curve of f and the x-axis is rotated \(360^\circ \) about the x-axis.

Find the volume of the solid formed.

Answer/Explanation

Markscheme

attempt to set up integral expression     M1

e.g. \(\pi {\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x\) , \(2\pi \int_0^2 {(16 – 4{x^2})} \) ,  \({\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x\)

\(\int {16} {\rm{d}}x = 16x\) , \(\int {4{x^2}{\rm{d}}x = } \frac{{4{x^3}}}{3}\) (seen anywhere)     A1A1

evidence of substituting limits into the integrand     (M1)

e.g. \(\left( {32 – \frac{{32}}{3}} \right) – \left( { – 32 + \frac{{32}}{3}} \right)\) , \(64 – \frac{{64}}{3}\)

volume \(= \frac{{128\pi }}{3}\)     A2     N3

[6 marks]

Question

Find \(\int_4^{10} {(x – 4){\rm{d}}x} \) .

[4]
a.

Part of the graph of \(f(x) = \sqrt {{x^{}} – 4} \) , for \(x \ge 4\) , is shown below. The shaded region R is enclosed by the graph of \(f\) , the line \(x = 10\) , and the x-axis.

The region R is rotated \({360^ \circ }\) about the x-axis. Find the volume of the solid formed.

[3]
b.
Answer/Explanation

Markscheme

correct integration     A1A1

e.g. \(\frac{{{x^2}}}{2} – 4x\), \(\left[ {\frac{{{x^2}}}{2} – 4x} \right]_4^{10}\), \(\frac{{{{(x – 4)}^2}}}{2}\)

Notes: In the first 2 examples, award A1 for each correct term.

In the third example, award A1 for \(\frac{1}{2}\) and A1 for \({(x – 4)^2}\).

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

e.g. \(\left( {\frac{{{{10}^2}}}{2} – 4(10)} \right) – \left( {\frac{{{4^2}}}{2} – 4(4)} \right),10 – ( – 8),\frac{1}{2}({6^2} – 0)\)

\(\int_4^{10} {(x – 4){\rm{d}}x = 18} \)     A1     N2

a.

attempt to substitute either limits or the function into volume formula     (M1)

e.g. \(\pi \int_4^{10} {{f^2}} {\rm{d}}x{\text{, }}{\int_a^b {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {\sqrt {x – 4} } \)

Note: Do not penalise for missing \(\pi \) or dx.

correct substitution (accept absence of dx and \(\pi \))     (A1)

e.g. \(\pi {\int_4^{10} {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {(x – 4){\rm{d}}x} {\text{, }}\int_4^{10} {(x – 4){\rm{d}}x} \)

volume = \(18\pi \)     A1     N2

[3 marks]

b.

Question

Let \(f(x) = {x^2}\).

Find \(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x} \).

[4]
a.

The following diagram shows part of the graph of \(f\).

 

The shaded region \(R\) is enclosed by the graph of \(f\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\).

Find the volume of the solid formed when \(R\) is revolved \({360^ \circ }\) about the \(x\)-axis.

[2]
b.
Answer/Explanation

Markscheme

substituting for \({\left( {f(x)} \right)^2}\) (may be seen in integral)     A1

eg     \({\left( {{x^2}} \right)^2}{\text{, }}{x^4}\)

correct integration, \(\int {{x^4}{\text{d}}x = \frac{1}{5}{x^5}} \)     (A1)

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

eg     \(\frac{{{2^5}}}{5} – \frac{1}{5}{\text{, }}\frac{1}{5}(1 – 4)\)

\(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x} = \frac{{31}}{5}( = 6.2) \)     A1     N2

[4 marks]

a.

attempt to substitute limits or function into formula involving \({f^2}\)     (M1)

eg     \(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x{\text{, }}\pi \int {{x^4}{\text{d}}x} } \)

\(\frac{{31}}{5}\pi {\text{   }}( = 6.2\pi )\)     A1     N2

[2 marks]

b.

Question

The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).

M17/5/MATME/SP1/ENG/TZ1/10

Show that \(\cos \theta  = \frac{3}{4}\).

[6]
a.

Given that \(\tan \theta  > 0\), find \(\tan \theta \).

[3]
b.

Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[6]
c.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(\sum {p = 1} \)

correct equation     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2\cos 2\theta  = 1\)

correct equation in \(\cos \theta \)     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2(2{\cos ^2}\theta  – 1) = 1,{\text{ }}4{\cos ^2}\theta  + \cos \theta  – 3 = 0\)

evidence of valid approach to solve quadratic     (M1)

eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}\)

correct working, clearly leading to required answer     A1

eg\(\,\,\,\,\,\)\((4\cos \theta  – 3)(\cos \theta  + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}\)

correct reason for rejecting \(\cos \theta  \ne  – 1\)     R1

eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta  > 0\)

Note:     Award R0 for \(\cos \theta  \ne  – 1\) without a reason.

\(\cos \theta  = \frac{3}{4}\)    AG  N0

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     

(A1)

eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)

\(\tan \theta  = \frac{{\sqrt 7 }}{3}\)     A1     N2

[3 marks]

b.

attempt to substitute either limits or the function into formula involving \({f^2}\)     (M1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)

correct substitution of both limits and function     (A1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\tan x\)

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

eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} – \tan \theta \)

Note:     Award M0 if they substitute into original or differentiated function.

\(\tan \frac{\pi }{4} = 1\)    (A1)

eg\(\,\,\,\,\,\)\(1 – \tan \theta \)

\(V = \pi  – \frac{{\pi \sqrt 7 }}{3}\)     A1     N3

[6 marks]

c.

Question

Let \(f\left( x \right) = \frac{1}{{\sqrt {2x – 1} }}\), for \(x > \frac{1}{2}\).

Find \(\int {{{\left( {f\left( x \right)} \right)}^2}{\text{d}}x} \).

[3]
a.

Part of the graph of f is shown in the following diagram.

The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.

[4]
b.
Answer/Explanation

Markscheme

correct working      (A1)

eg   \(\int {\frac{1}{{2x – 1}}{\text{d}}x,\,\,\int {{{\left( {2x – 1} \right)}^{ – 1}},\,\,\frac{1}{{2x – 1}},\,\,\int {{{\left( {\frac{1}{{\sqrt u }}} \right)}^2}\frac{{{\text{d}}u}}{2}} } } \)

\({\int {\left( {f\left( x \right)} \right)} ^2}{\text{d}}x = \frac{1}{2}{\text{ln}}\left( {2x – 1} \right) + c\)      A2 N3

Note: Award A1 for \(\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)\).

[3 marks]

a.

attempt to substitute either limits or the function into formula involving f 2 (accept absence of \(\pi \) / dx)     (M1)

eg   \(\int_1^9 {{y^2}{\text{d}}x,\,\,} \pi {\int {\left( {\frac{1}{{\sqrt {2x – 1} }}} \right)} ^2}{\text{d}}x,\,\,\left[ {\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)} \right]_1^9\)

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

eg  \(\frac{\pi }{2}\left( {{\text{ln}}\left( {17} \right) – {\text{ln}}\left( 1 \right)} \right),\,\,\pi \left( {0 – \frac{1}{2}{\text{ln}}\left( {2 \times 9 – 1} \right)} \right)\)

correct working involving calculating a log value or using log law     (A1)

eg  \({\text{ln}}\left( 1 \right) = 0,\,\,{\text{ln}}\left( {\frac{{17}}{1}} \right)\)

\(\frac{\pi }{2}{\text{ln}}17\,\,\,\,\left( {{\text{accept }}\pi {\text{ln}}\sqrt {17} } \right)\)    A1 N3

Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.

[4 marks]

b.
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