1.08d Evaluate definite integrals: between limits

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with polar equation $$r = a + 3 \cos \theta , \quad a > 0 , \quad 0 \leqslant \theta < 2 \pi$$ The area enclosed by the curve is \(\frac { 107 } { 2 } \pi\).
Find the value of \(a\).

5.

1. Use de Moivre's theorem to show that $$\sin ^ { 5 } \theta \equiv a \sin 5 \theta + b \sin 3 \theta + c \sin \theta$$ where \(a\), \(b\) and \(c\) are constants to be found.
2. Hence show that \(\int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 5 } \theta \mathrm {~d} \theta = \frac { 53 } { 480 }\) VILM SIHI NITIIIUMI ON OC
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8. The curve \(C\) has parametric equations $$x = \theta - \sin \theta , \quad y = 1 - \cos \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$ The curve \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis. The area of the curved surface generated is given by \(S\).

1. Show that $$S = 2 \pi \sqrt { 2 } \int _ { 0 } ^ { 2 \pi } ( 1 - \cos \theta ) ^ { \frac { 3 } { 2 } } \mathrm {~d} \theta$$
2. Hence find the exact value of \(S\).

8. $$I _ { n } = \int \frac { x ^ { n } } { \sqrt { \left( x ^ { 2 } + k ^ { 2 } \right) } } \mathrm { d } x \quad \text { where } k \text { is a constant and } n \in \mathbb { Z } ^ { + }$$

1. Show that, for \(n \geqslant 2\) $$I _ { n } = \frac { x ^ { n - 1 } } { n } \left( x ^ { 2 } + k ^ { 2 } \right) ^ { \frac { 1 } { 2 } } - \frac { ( n - 1 ) } { n } k ^ { 2 } I _ { n - 2 }$$
2. Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { \sqrt { \left( x ^ { 2 } + 1 \right) } } \mathrm { d } x$$

Use calculus to find the exact value of \(\int _ { 1 } ^ { 2 } \left( 3 x ^ { 2 } + 5 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x\).

\begin{figure}[h] \end{figure} Figure 1 shows part of the curve \(y = \frac { 3 } { \sqrt { } ( 1 + 4 x ) }\). The region \(R\) is bounded by the curve, the \(x\)-axis, and the lines \(x = 0\) and \(x = 2\), as shown shaded in Figure 1.

1. Use integration to find the area of \(R\). The region \(R\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis.
2. Use integration to find the exact value of the volume of the solid formed.

5. A particle moves along the \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0\), the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\) in the direction of \(x\) increasing, where \(v = 2 t ^ { \frac { 3 } { 2 } } - 6 t + 2\) At time \(t = 0\) the particle passes through the origin \(O\). At the instant when the acceleration of the particle is zero, the particle is at the point \(A\). Find the distance \(O A\).
(8)

2.

1. Find $$\int \frac { 1 } { \sqrt { } \left( 4 x ^ { 2 } + 9 \right) } d x$$
2. Use your answer to part (a) to find the exact value of $$\int _ { - 3 } ^ { 3 } \frac { 1 } { \sqrt { \left( 4 x ^ { 2 } + 9 \right) } } d x$$ giving your answer in the form \(k \ln ( a + b \sqrt { } 5 )\), where \(a\) and \(b\) are integers and \(k\) is a constant.
   

3 For a restaurant with a home-delivery service, the delivery time in minutes can be modelled by a continuous random variable \(T\) with probability density function given by $$f ( t ) = \begin{cases} \frac { \pi } { 90 } \sin \left( \frac { \pi t } { 60 } \right) & 20 \leqslant t \leqslant 60 \\ 0 & \text { otherwise. } \end{cases}$$

1. Given that \(20 \leqslant a \leqslant 60\), show that \(\mathrm { P } ( T \leqslant a ) = \frac { 1 } { 3 } \left( 1 - 2 \cos \left( \frac { \pi a } { 60 } \right) \right)\). There is a delivery charge of \(\pounds 3\) but this is reduced to \(\pounds 2\) if the delivery time exceeds a minutes.
2. Find the value of \(a\) for which the expected value of the delivery charge for a home-delivery is £2.80.

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\).

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

6

1. Find \(\int \left( x ^ { \frac { 1 } { 2 } } + 4 \right) \mathrm { d } x\).
2. 1. Find the value, in terms of \(a\), of \(\int _ { 1 } ^ { a } 4 x ^ { - 2 } \mathrm {~d} x\), where \(a\) is a constant greater than 1 .
   2. Deduce the value of \(\int _ { 1 } ^ { \infty } 4 x ^ { - 2 } \mathrm {~d} x\).

8 The cubic polynomial \(2 x ^ { 3 } + k x ^ { 2 } - x + 6\) is denoted by \(\mathrm { f } ( x )\). It is given that \(( x + 1 )\) is a factor of \(\mathrm { f } ( x )\).

1. Show that \(k = - 5\), and factorise \(\mathrm { f } ( x )\) completely.
2. Find \(\int _ { - 1 } ^ { 2 } f ( x ) \mathrm { d } x\).
3. Explain with the aid of a sketch why the answer to part (ii) does not give the area of the region between the curve \(y = \mathrm { f } ( x )\) and the \(x\)-axis for \(- 1 \leqslant x \leqslant 2\). \section*{[Question 9 is printed overleaf.]}

6

1. 1. Find \(\int x \left( x ^ { 2 } - 4 \right) d x\)
   2. Hence evaluate \(\int _ { 1 } ^ { 6 } x \left( x ^ { 2 } - 4 \right) d x\).
2. Find \(\int \frac { 6 } { x ^ { 3 } } d x\)

9

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_506_812_676_653} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

10 The equation of a curve is \(y = 7 + 6 x - x ^ { 2 }\).

1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
3. The curve and the line \(y = 12\) intersect at ( 1,12 ) and ( 5,12 ). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{15b8f97b-c058-409f-907f-cb0a6102abc4-5_643_1034_331_513} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

10 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10-second intervals. \begin{figure}[h] \end{figure} The measured speeds are shown below.

1. Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line \(t = 60\) and the axes. [This area represents the distance travelled by the car.]
2. Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by \(v = 28 - t + 0.015 t ^ { 2 }\).
3. Show that the difference between the value given by the model when \(t = 10\) and the measured value is less than \(3 \%\) of the measured value.
4. According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 - t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

2 Find \(\int _ { 1 } ^ { 2 } \left( 12 x ^ { 5 } + 5 \right) \mathrm { d } x\).

6 Evaluate \(\int _ { 1 } ^ { 2 } \left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x\).

2. \includegraphics[max width=\textwidth, alt={}, center]{5025c118-e763-424b-b2c1-5452953a43a9-1_550_901_817_468} The diagram shows the curve with equation \(y = \sqrt { x } + \frac { 8 } { x ^ { 2 } } , x > 0\).
Show that the area of the shaded region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 9\) is \(24 \frac { 4 } { 9 }\).

6. Evaluate

1. \(\quad \int _ { 1 } ^ { 4 } \left( x ^ { 2 } - 5 x + 4 \right) \mathrm { d } x\),
2. \(\int _ { - \infty } ^ { - 1 } \frac { 1 } { x ^ { 4 } } \mathrm {~d} x\).

8. The finite region \(R\) is bounded by the curve \(y = 1 + 3 \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 8\).

1. Use the trapezium rule with three intervals, each of width 2 , to estimate to 3 significant figures the area of \(R\).
2. Use integration to find the exact area of \(R\) in the form \(a + b \sqrt { 2 }\).
3. Find the percentage error in the estimate made in part (a).

6.

1. Evaluate $$\int _ { 2 } ^ { 4 } \left( 2 - \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x$$
2. Given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { 3 } + 1$$ and that \(y = 3\) when \(x = 0\), find the value of \(y\) when \(x = 2\).

4 A cyclist travels along a straight road. Her velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at time \(t\) seconds after starting from a point \(O\), is given by $$\begin{aligned} & v = 2 \quad \text { for } 0 \leqslant t \leqslant 10 \\ & v = 0.03 t ^ { 2 } - 0.3 t + 2 \quad \text { for } t \geqslant 10 . \end{aligned}$$

1. Find the displacement of the cyclist from \(O\) when \(t = 10\).
2. Show that, for \(t \geqslant 10\), the displacement of the cyclist from \(O\) is given by the expression \(0.01 t ^ { 3 } - 0.15 t ^ { 2 } + 2 t + 5\).
3. Find the time when the acceleration of the cyclist is \(0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Hence find the displacement of the cyclist from \(O\) when her acceleration is \(0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).