1.08d Evaluate definite integrals: between limits

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Edexcel FM2 2024 June Q4
12 marks Standard +0.8
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c14975b7-6afa-44ce-beab-1cba2e82b249-14_675_528_242_772} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} A uniform lamina \(O A B\) is in the shape of the region \(R\).
Region \(R\) lies in the first quadrant and is bounded by the curve with equation \(\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 36 } = 1\), the \(x\)-axis, and the \(y\)-axis, as shown shaded in Figure 3. The point \(A\) is the point of intersection of the curve and the \(x\)-axis.
The point \(B\) is the point of intersection of the curve and the \(y\)-axis.
One unit on each axis represents 1 m .
The area of \(R\) is \(6 \pi\) The centre of mass of \(R\) lies at the point with coordinates \(( \bar { x } , \bar { y } )\)
  1. Use algebraic integration to show that \(\bar { x } = \frac { 16 } { 3 \pi }\)
  2. Use algebraic integration to find the exact value of \(\bar { y }\) The lamina is freely suspended from \(A\) and hangs in equilibrium with \(O A\) at angle \(\theta ^ { \circ }\) to the downward vertical.
  3. Find the value of \(\theta\)
Edexcel FM2 Specimen Q1
7 marks Standard +0.3
  1. A flag pole is 15 m long.
The flag pole is non-uniform so that, at a distance \(x\) metres from its base, the mass per unit length of the flag pole, \(m \mathrm {~kg} \mathrm {~m} ^ { - 1 }\) is given by the formula \(m = 10 \left( 1 - \frac { x } { 25 } \right)\). The flag pole is modelled as a rod.
  1. Show that the mass of the flag pole is 105 kg .
  2. Find the distance of the centre of mass of the flag pole from its base.
CAIE P1 2022 November Q10
10 marks Standard +0.3
  1. Find the coordinates of \(A\).
  2. Find the volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Give your answer in the form \(\frac { \pi } { a } ( b \sqrt { c } - d )\), where \(a , b , c\) and \(d\) are integers.
  3. Find an exact expression for the perimeter of the shaded region.
CAIE P2 2024 November Q6
9 marks Moderate -0.3
  1. Use the trapezium rule with two intervals to find an approximation to the area of region \(A\). Give your answer correct to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-10_2720_38_105_2010} \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-11_2716_29_107_22}
  2. Find the exact total area of regions \(A\) and \(B\). Give your answer in the form \(k \ln m\), where \(k\) and \(m\) are constants.
  3. Deduce an approximation to the area of region \(B\). Give your answer correct to 3 significant figures.
  4. State, with a reason, whether your answer to part (c) is an over-estimate or an under-estimate of the area of region \(B\).
CAIE P2 2007 November Q7
8 marks Standard +0.3
  1. Prove the identity $$( \cos x + 3 \sin x ) ^ { 2 } \equiv 5 - 4 \cos 2 x + 3 \sin 2 x$$
  2. Using the identity, or otherwise, find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( \cos x + 3 \sin x ) ^ { 2 } d x$$
CAIE P3 2020 June Q7
9 marks Challenging +1.2
  1. Express \(\mathrm { f } ( x )\) in partial fractions.
  2. Using your answer to part (a), show that $$( f ( x ) ) ^ { 2 } = \frac { 1 } { ( 2 x - 1 ) ^ { 2 } } - \frac { 1 } { 2 x - 1 } + \frac { 1 } { 2 x + 1 } + \frac { 1 } { ( 2 x + 1 ) ^ { 2 } }$$
  3. Hence show that \(\int _ { 1 } ^ { 2 } ( \mathrm { f } ( x ) ) ^ { 2 } \mathrm {~d} x = \frac { 2 } { 5 } + \frac { 1 } { 2 } \ln \left( \frac { 5 } { 9 } \right)\).
CAIE P3 2023 June Q10
10 marks Standard +0.3
  1. Find the exact coordinates of \(M\).
  2. Using the substitution \(u = 3 - 2 x\), find by integration the area of the shaded region bounded by the curve and the \(x\)-axis. Give your answer in the form \(a \sqrt { 13 }\), where \(a\) is a rational number. [5]
CAIE P3 2021 November Q9
10 marks Standard +0.3
  1. Find the \(x\)-coordinate of the stationary point of the curve with equation \(y = \mathrm { f } ( x )\).
  2. Using the substitution \(u = \sqrt { x }\), show that \(\int _ { 0 } ^ { 4 } \mathrm { f } ( x ) \mathrm { d } x = \frac { 1 } { 3 } \ln 5\).
CAIE P3 2024 November Q6
8 marks Standard +0.3
  1. Given that the \(x\)-coordinate of \(M\) lies in the interval \(\frac { 1 } { 2 } \pi < x < \frac { 3 } { 4 } \pi\), find the exact coordinates of \(M\). \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-10_2718_35_107_2012} \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-11_2725_35_99_20}
  2. Find the exact area of the region \(R\).
OCR MEI C2 2011 January Q11
11 marks Moderate -0.3
  1. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
  2. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.
OCR MEI C2 Q11
12 marks Moderate -0.8
  1. The speed-time graph on the insert sheet provides the axes and the first two points plotted. Plot the remainder of these points and join them with a smooth curve. The area between this curve and the \(t\)-axis represents the distance travelled by the car in this time.
  2. Using the trapezium rule with 6 values of \(t\) estimate the area under the curve to give the distance travelled. Illustrate on your graph the area found.
  3. John's teacher suggests that the equation of the curve could be \(v = 6 t - \frac { 1 } { 2 } t ^ { 2 }\). Find, by calculus, the area between this curve and the \(t\) axis.
  4. Plot this curve on your graph. Comment on whether the estimates obtained in parts (ii) and (iii) are overestimates or underestimates. 12 Fig. 12 shows a window. The base and sides are parts of a rectangle with dimensions \(2 x\) metres horizontally by \(y\) metres vertically. The top is a semicircle of radius \(x\) metres. The perimeter of the window is 10 metres. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{73d1c02b-1b7b-426d-a171-c762597cfed4-4_428_433_1638_766} \captionsetup{labelformat=empty} \caption{Fig. 12}
    \end{figure}
  1. Express \(y\) as a function of \(x\).
  2. Find the total area, \(A \mathrm {~m} ^ { 2 }\), in terms of \(x\) and \(y\). Use your answer to part (i) to show that this simplifies to $$A = 10 x - 2 x ^ { 2 } - \frac { 1 } { 2 } \pi x ^ { 2 }$$
  3. Prove that for the maximum value of \(A\), \(y = x\) exactly.
    \section*{MEI STRUCTURED MATHEMATICS } \section*{CONCEPTS FOR ADVANCED MATHEMATICS, C2} \section*{Practice Paper C2-B
    Insert sheet for question 11}
Edexcel AEA 2007 June Q6
17 marks Hard +2.3
  1. Find an expression, in terms of \(x\), for the area \(A\) of \(R\).
  2. Show that \(\frac { \mathrm { d } A } { \mathrm {~d} x } = \frac { 1 } { 4 } ( \pi - 2 x - 2 \sin x ) \sec ^ { 2 } \frac { x } { 2 }\).
  3. Prove that the maximum value of \(A\) occurs when \(\frac { \pi } { 4 } < x < \frac { \pi } { 3 }\).
  4. Prove that \(\tan \frac { \pi } { 8 } = \sqrt { } 2 - 1\).
  5. Show that the maximum value of \(A > \frac { \pi } { 4 } ( \sqrt { } 2 - 1 )\).
OCR PURE Q8
9 marks Standard +0.3
8 In this question you must show detailed reasoning. The diagram shows part of the graph of \(y = 2 x ^ { \frac { 1 } { 3 } } - \frac { 7 } { x ^ { \frac { 1 } { 3 } } }\). The shaded region is enclosed by the curve, the \(x\)-axis and the lines \(x = 8\) and \(x = a\), where \(a > 8\). \includegraphics[max width=\textwidth, alt={}, center]{efde7b10-b4f3-469f-ba91-b765a16ea835-5_577_1164_477_438} Given that the area of the shaded region is 45 square units, find the value of \(a\).
OCR PURE Q8
7 marks Moderate -0.3
8 In this question you must show detailed reasoning. Given that \(\int _ { 4 } ^ { a } \left( \frac { 4 } { \sqrt { x } } + 3 \right) \mathrm { d } x = 7\), find the value of \(a\).
OCR MEI Paper 2 2020 November Q10
9 marks Standard +0.3
10 In this question you must show detailed reasoning. The equation of a curve is $$y = \frac { \sin 2 x - x } { x \sin x }$$
  1. Use the small angle approximation given in the list of formulae on pages 2-3 of this question paper to show that $$\int _ { 0.01 } ^ { 0.05 } \mathrm { ydx } \approx \ln 5$$
  2. Use the same small angle approximation to show that $$\frac { d y } { d x } \approx - 10000 \text { at the point where } x = 0.01 \text {. }$$ The equation \(y = 0\) has a root near \(x = 1\). Joan uses the Newton-Raphson method to find this root. The output from the spreadsheet she uses is shown in Fig. 10.1. \begin{table}[h]
    \(n\)01234567
    \(\mathrm { x } _ { \mathrm { n } }\)10.9585090.9500840.9482610.947860.9477720.9477530.947748
    \captionsetup{labelformat=empty} \caption{Fig. 10.1}
    \end{table} Joan carries out some analysis of this output. The results are shown in Fig. 10.2. \begin{table}[h]
    \(x\)\(y\)
    0.9477475\(- 7.79967 \mathrm { E } - 07\)
    0.9477485\(- 2.90821 \mathrm { E } - 06\)
    \(x\)\(y\)
    0.947745\(4.54066 \mathrm { E } - 06\)
    0.947755\(- 1.67417 \mathrm { E } - 05\)
    \captionsetup{labelformat=empty} \caption{Fig. 10.2}
    \end{table}
  3. Consider the information in Fig. 10.1 and Fig. 10.2.
OCR MEI Paper 2 2020 November Q14
8 marks Challenging +1.2
14 In this question you must show detailed reasoning. Fig. 14 shows the graphs of \(y = \sin x \cos 2 x\) and \(y = \frac { 1 } { 2 } - \sin 2 x \cos x\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cea67565-8074-4703-8e1a-09b98e380baf-16_647_898_404_233} \captionsetup{labelformat=empty} \caption{Fig. 14}
\end{figure} Use integration to find the area between the two curves, giving your answer in an exact form.
OCR MEI Paper 3 2024 June Q3
4 marks Standard +0.8
3 In this question you must show detailed reasoning. The diagram shows the curve with equation \(y = x ^ { 5 }\) and the square \(O A B C\) where the points \(A , B\) and \(C\) have coordinates \(( 1,0 ) , ( 1,1 )\) and \(( 0,1 )\) respectively. The curve cuts the square into two parts. \includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-04_658_780_1318_230} Show that the relationship between the areas of the two parts of the square is \(\frac { \text { Area to left of curve } } { \text { Area below curve } } = 5\).
OCR MEI Paper 3 Specimen Q5
5 marks Standard +0.3
5 In this question you must show detailed reasoning. Fig. 5 shows the circle with equation \(( x - 4 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 10\).
The points \(( 1,0 )\) and \(( 7,0 )\) lie on the circle. The point C is the centre of the circle. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b4e10fd2-4144-4019-bf00-070f93a2b05d-05_878_1000_685_255} \captionsetup{labelformat=empty} \caption{Fig. 5}
\end{figure} Find the area of the part of the circle below the \(x\)-axis.
OCR MEI Paper 1 2019 June Q1
3 marks Easy -1.2
1 In this question you must show detailed reasoning. Show that \(\int _ { 4 } ^ { 9 } ( 2 x + \sqrt { x } ) \mathrm { d } x = \frac { 233 } { 3 }\).
OCR H240/01 2018 September Q8
9 marks Standard +0.3
8 \includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-6_533_524_246_772} The diagram shows a container which consists of a cylinder with a solid base and a hemispherical top. The radius of the cylinder is \(r \mathrm {~cm}\) and the height is \(h \mathrm {~cm}\). The container is to be made of thin plastic. The volume of the container is \(45 \pi \mathrm {~cm} ^ { 3 }\).
  1. Show that the surface area of the container, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = \frac { 5 } { 3 } \pi r ^ { 2 } + \frac { 90 \pi } { r } .$$ [The volume of a sphere is \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and the surface area of a sphere is \(S = 4 \pi r ^ { 2 }\).]
  2. Use calculus to find the minimum surface area of the container, justifying that it is a minimum.
  3. Suggest a reason why the manufacturer would wish to minimise the surface area.
AQA C2 2007 June Q1
8 marks Easy -1.3
1
  1. Simplify:
    1. \(x ^ { \frac { 3 } { 2 } } \times x ^ { \frac { 1 } { 2 } }\);
    2. \(x ^ { \frac { 3 } { 2 } } \div x\);
    3. \(\left( x ^ { \frac { 3 } { 2 } } \right) ^ { 2 }\).
    1. Find \(\int 3 x ^ { \frac { 1 } { 2 } } \mathrm {~d} x\).
    2. Hence find the value of \(\int _ { 1 } ^ { 9 } 3 x ^ { \frac { 1 } { 2 } } \mathrm {~d} x\).
OCR C2 Q5
8 marks Moderate -0.8
5. The curve \(y = \mathrm { f } ( x )\) passes through the point \(P ( - 1,3 )\) and is such that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 } { x ^ { 3 } } , \quad x \neq 0$$
  1. Find \(\mathrm { f } ( x )\).
  2. Show that the area of the finite region bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\) is \(4 \frac { 1 } { 2 }\).
OCR C2 Q8
11 marks Standard +0.3
  1. Given that $$\int _ { 1 } ^ { 3 } \left( x ^ { 2 } - 2 x + k \right) d x = 8 \frac { 2 } { 3 }$$ find the value of the constant \(k\).
  2. Evaluate $$\int _ { 2 } ^ { \infty } \frac { 6 } { x ^ { \frac { 5 } { 2 } } } \mathrm {~d} x$$ giving your answer in its simplest form.
AQA C3 2007 June Q8
12 marks Standard +0.3
8
  1. Write down \(\int \sec ^ { 2 } x \mathrm {~d} x\).
  2. Given that \(y = \frac { \cos x } { \sin x }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
  3. Prove the identity \(( \tan x + \cot x ) ^ { 2 } = \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x\).
  4. Hence find \(\int _ { 0.5 } ^ { 1 } ( \tan x + \cot x ) ^ { 2 } \mathrm {~d} x\), giving your answer to two significant figures.
OCR MEI C4 2006 January Q6
7 marks Standard +0.3
6
  1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 - x ^ { 2 } } }\) for \(| x | < 2\).
  2. Use this result to find an approximation for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to
    4 significant figures.
  3. Given that \(\int \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c\), evaluate \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to 4 significant figures.