1.08d Evaluate definite integrals: between limits

6 A particle starts from rest at the point A and travels in a straight line. The displacement sm of the particle from A at time ts after leaving A is given by $$s = 0.001 t ^ { 4 } - 0.04 t ^ { 3 } + 0.6 t ^ { 2 } , \quad \text { for } 0 \leqslant t \leqslant 10 .$$

1. Show that the velocity of the particle is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when \(\mathrm { t } = 10\). The acceleration of the particle for \(t \geqslant 10\) is \(( 0.8 - 0.08 t ) \mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Show that the velocity of the particle is zero when \(\mathrm { t } = 20\).
3. Find the displacement from A of the particle when \(\mathrm { t } = 20\).

2 A particle moves along the \(x\)-axis with velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at time \(t\) given by $$v = 24 t - 6 t ^ { 2 }$$ The positive direction is in the sense of \(x\) increasing.

1. Find an expression for the acceleration of the particle at time \(t\).
2. Find the times, \(t _ { 1 }\) and \(t _ { 2 }\), at which the particle has zero speed.
3. Find the distance travelled between the times \(t _ { 1 }\) and \(t _ { 2 }\).

6 A toy car is travelling in a straight horizontal line.
One model of the motion for \(0 \leqslant t \leqslant 8\), where \(t\) is the time in seconds, is shown in the velocity-time graph Fig. 6. \begin{figure}[h] \end{figure}

1. Calculate the distance travelled by the car from \(t = 0\) to \(t = 8\).
2. How much less time would the car have taken to travel this distance if it had maintained its initial speed throughout?
3. What is the acceleration of the car when \(t = 1\) ? From \(t = 8\) to \(t = 14\), the car travels 58.5 m with a new constant acceleration, \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
4. Find \(a\). A second model for the velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of the toy car is $$v = 12 - 10 t + \frac { 9 } { 4 } t ^ { 2 } - \frac { 1 } { 8 } t ^ { 3 } , \text { for } 0 \leqslant t \leqslant 8$$ This model agrees with the values for \(v\) given in Fig. 6 for \(t = 0,2,4\) and 6. [Note that you are not required to verify this.] Use this second model to answer the following questions.
5. Calculate the acceleration of the car when \(t = 1\).
6. Initially the car is at A. Find an expression in terms of \(t\) for the displacement of the car from A after the first \(t\) seconds of its motion. Hence find the displacement of the car from A when \(t = 8\).
7. Explain with a reason what this model predicts for the motion of the car between \(t = 2\) and \(t = 4\).

2. The diagram shows the curve with equation \(y = \frac { 3 x + 1 } { \sqrt { x } } , x > 0\).
The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
Find the volume of the solid formed when the shaded region is rotated through four right angles about the \(x\)-axis, giving your answer in the form \(\pi ( a + \ln b )\), where \(a\) and \(b\) are integers.

7 Fig. 7 is a sketch of part of the velocity-time graph for the motion of an insect walking in a straight line. Its velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at time \(t\) seconds for the time interval \(- 3 \leqslant t \leqslant 5\) is given by $$v = t ^ { 2 } - 2 t - 8 .$$ \begin{figure}[h] \end{figure}

1. Write down the velocity of the insect when \(t = 0\).
2. Show that the insect is instantaneously at rest when \(t = - 2\) and when \(t = 4\).
3. Determine the velocity of the insect when its acceleration is zero. Write down the coordinates of the point A shown in Fig. 7.
4. Calculate the distance travelled by the insect from \(t = 1\) to \(t = 4\).
5. Write down the distance travelled by the insect in the time interval \(- 2 \leqslant t \leqslant 4\).
6. How far does the insect walk in the time interval \(1 \leqslant t \leqslant 5\) ?

8 The displacement, \(x \mathrm {~m}\), from the origin O of a particle on the \(x\)-axis is given by $$x = 10 + 36 t + 3 t ^ { 2 } - 2 t ^ { 3 }$$ where \(t\) is the time in seconds and \(- 4 \leqslant t \leqslant 6\).

1. Write down the displacement of the particle when \(t = 0\).
2. Find an expression in terms of \(t\) for the velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of the particle.
3. Find an expression in terms of \(t\) for the acceleration of the particle.
4. Find the maximum value of \(v\) in the interval \(- 4 \leqslant t \leqslant 6\).
5. Show that \(v = 0\) only when \(t = - 2\) and when \(t = 3\). Find the values of \(x\) at these times.
6. Calculate the distance travelled by the particle from \(t = 0\) to \(t = 4\).
7. Determine how many times the particle passes through O in the interval \(- 4 \leqslant t \leqslant 6\).

3 Fig. 3 is a sketch of the velocity-time graph modelling the velocity of a sprinter at the start of a race. \begin{figure}[h] \end{figure}

1. How can you tell from the sketch that the acceleration is not modelled as being constant for \(0 \leqslant t \leqslant 4\) ? The velocity of the sprinter, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), for the time interval \(0 \leqslant t \leqslant 4\) is modelled by the expression $$v = 3 t - \frac { 3 } { 8 } t ^ { 2 }$$
2. Find the acceleration that the model predicts for \(t = 4\) and comment on what this suggests about the running of the sprinter.
3. Calculate the distance run by the sprinter from \(t = 1\) to \(t = 4\).

5. The diagram shows the curve with equation \(y = \frac { 1 } { \sqrt { 3 x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 5\).

1. Find the area of the shaded region. The shaded region is rotated through four right angles about the \(x\)-axis.
2. Find the volume of the solid formed, giving your answer in the form \(k \pi \ln 2\).

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

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 \(y = 12\).

1 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 \(y = 12\).

4. The finite region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\).

1. Find the exact area of \(R\).
2. Show that the volume of the solid formed when \(R\) is rotated through four right angles about the \(x\)-axis is \(\frac { 1 } { 3 } \pi\).

4 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the equation of the tangent to the curve at the point on the curve where \(x = 2\). Show that this tangent crosses the \(x\)-axis where \(x = \frac { 2 } { 3 }\).
2. Show that the curve crosses the \(x\)-axis where \(x = 1\) and find the \(x\)-coordinate of the other point of intersection of the curve with the \(x\)-axis.
3. Find \(\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x\). Hence find the area of the region bounded by the curve, the tangent and the \(x\)-axis, shown shaded on Fig. 10. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{12e190fc-437f-499d-9c27-da49a7546755-3_643_1034_267_549} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find, in exact form, the range of values of \(x\) for which \(x ^ { 3 } - 6 x + 2\) is a decreasing function.
3. Find the equation of the tangent to the curve at the point \(( - 1,7 )\). Find also the coordinates of the point where this tangent crosses the curve again.

1 Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for \(x\) and \(y\) are metres. \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 5 strips to estimate the area of the end wall of the building.
2. Oskar now uses the equation \(y = - 0.001 x ^ { 3 } - 0.025 x ^ { 2 } + 0.6 x + 9\), for \(0 \leqslant x \leqslant 15\), to model the curve of the roof.
   (A) Calculate the difference between the height of the roof when \(x = 12\) given by this model and the data shown in Fig. 12.
   (B) Use integration to find the area of the end wall given by this model.

3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

6. Find the value of each of the following integrals in exact, simplified form.

1. \(\quad \int _ { - 1 } ^ { 0 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x\)
2. \(\int _ { 2 } ^ { 4 } \frac { 3 x ^ { 2 } - 2 } { x } \mathrm {~d} x\)

1 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 $$\int _ { 0 } ^ { 60 } \left( 28 \quad t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O . \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. The curve of the roof may be modelled by \(y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.
3. Use integration to find the area of the cross-section according to this model.
4. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\).

5 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the equation of the tangent to the curve at the point on the curve where \(x = 2\). Show that this tangent crosses the \(x\)-axis where \(x = \frac { 2 } { 3 }\).
2. Show that the curve crosses the \(x\)-axis where \(x = 1\) and find the \(x\)-coordinate of the other point of intersection of the curve with the \(x\)-axis.
3. Find \(\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x\). Hence find the area of the region bounded by the curve, the tangent and the \(x\)-axis, shown shaded on Fig. 10.

3

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]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700} \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.
   [0pt] [5]
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]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629} \captionsetup{labelformat=empty} \caption{Not to scale}
   \end{figure} Fig. 9.2 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.

4. \includegraphics[max width=\textwidth, alt={}, center]{c0b79c3c-9537-4c71-903b-01434dfb5d26-1_492_803_1562_452} The diagram shows the curves \(y = ( x - 1 ) ^ { 2 }\) and \(y = 2 - \frac { 2 } { x } , x > 0\).

1. Verify that the two curves meet at the points where \(x = 1\) and where \(x = 2\). The shaded region bounded by the two curves is rotated completely about the \(x\)-axis.
2. Find the exact volume of the solid formed.

5. The finite region \(R\) is bounded by the curve with equation \(y = \sqrt [ 3 ] { 3 x - 1 }\), the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 3\).

1. Find the area of \(R\).
2. Find, in terms of \(\pi\), the volume of the solid formed when \(R\) is rotated through four right angles about the \(x\)-axis.

1 Show that \(\int _ { 2 } ^ { 8 } \frac { 3 } { x } \mathrm {~d} x = \ln 64\).

7

1. Find the exact value of \(\int _ { 1 } ^ { 2 } \frac { 2 } { ( 4 x - 1 ) ^ { 2 } } \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{ebfdf170-99c6-4785-b9d7-201c3425b4c9-3_563_753_1681_735} The diagram shows part of the curve \(y = \frac { 1 } { x }\). The point \(P\) has coordinates \(\left( a , \frac { 1 } { a } \right)\) and the point \(Q\) has coordinates \(\left( 2 a , \frac { 1 } { 2 a } \right)\), where \(a\) is a positive constant. The point \(R\) is such that \(P R\) is parallel to the \(x\)-axis and \(Q R\) is parallel to the \(y\)-axis. The region shaded in the diagram is bounded by the curve and by the lines \(P R\) and \(Q R\). Show that the area of this shaded region is \(\ln \left( \frac { 1 } { 2 } \mathrm { e } \right)\).

4 The integral I is defined by $$I = \int _ { 0 } ^ { 13 } ( 2 x + 1 ) ^ { \frac { 1 } { 3 } } d x$$

1. Use integration to find the exact value of I .
2. Use Simpson's rule with two strips to find an approximate value for I. Give your answer correct to 3 significant figures.

6

1. Given that \(\int _ { 0 } ^ { \mathrm { a } } \left( 6 \mathrm { e } ^ { 2 \mathrm { x } } + \mathrm { x } \right) \mathrm { dx } = 42\), show that \(\mathrm { a } = \frac { 1 } { 2 } \ln \left( 15 - \frac { 1 } { 6 } \mathrm { a } ^ { 2 } \right)\).
2. Use an iterative formula, based on the equation in part (i), to find the value of a correct to 3 decimal places. Use a starting value of 1 and show the result of each iteration.