1.08d Evaluate definite integrals: between limits

8

1. The quadratic polynomial \(a x ^ { 2 } + b x\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { f } ( x )\).
   Use differentiation from first principles to determine, in terms of \(a , b\) and \(x\), an expression for \(\mathrm { f } ^ { \prime } ( x )\).
2. \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-07_565_1043_516_317} $$y = a x ^ { 2 } + b x$$ The diagram shows the quadratic curve \(y = a x ^ { 2 } + b x\), where \(a\) and \(b\) are constants. The shaded region is enclosed by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\). The tangent to the curve at \(x = 4\) intersects the \(x\)-axis at the point with coordinates \(( k , 0 )\).
   Given that the area of the shaded region is 9 units \({ } ^ { 2 }\), and the gradient of this tangent is \(- \frac { 3 } { 4 }\), determine the value of \(k\).

11 \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-08_586_672_1231_242} A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\) in the positive \(x\)-direction is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that \(v = t ( t - 3 ) ( 8 - t )\). \(P\) attains its maximum velocity at time \(T\) seconds. The diagram shows part of the velocity-time graph for the motion of \(P\).

1. State the acceleration of \(P\) at time \(T\).
2. In this question you must show detailed reasoning. Determine the value of \(T\).
3. Find the total distance that \(P\) travels between times \(t = 0\) and \(t = T\). \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-09_524_410_251_242} Particles \(P\) and \(Q\), of masses 4 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The system is in equilibrium with \(P\) hanging 1.75 m above a horizontal plane and \(Q\) resting on the plane. Both parts of the string below the pulley are vertical (see diagram).
   1. Find the magnitude of the normal reaction force acting on \(Q\). The mass of \(P\) is doubled, and the system is released from rest. You may assume that in the subsequent motion \(Q\) does not reach the pulley.
   2. Determine the magnitude of the force exerted on the pulley by the string before \(P\) strikes the plane.
   3. Determine the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest. When this motion is observed in practice, it is found that the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest is less than the answer calculated in part (c).
   4. State one factor that could account for this difference.

10 A particle \(P\) is moving in a straight line. At time \(t\) seconds, where \(t \geqslant 0 , P\) has velocity \(v \mathrm {~ms} ^ { - 1 }\) and acceleration \(a \mathrm {~ms} ^ { - 2 }\) where \(a = 4 t - 9\). It is given that \(v = 2\) when \(t = 1\).

1. Find an expression for \(v\) in terms of \(t\). The particle \(P\) is instantaneously at rest when \(t = t _ { 1 }\) and \(t = t _ { 2 }\), where \(t _ { 1 } < t _ { 2 }\).
2. Find the values of \(t _ { 1 }\) and \(t _ { 2 }\).
3. Determine the total distance travelled by \(P\) between times \(t = 0\) and \(t = t _ { 2 }\).

6 In this question you must show detailed reasoning.
A particle moves in a straight line. Its velocity \(v \mathrm {~ms} ^ { - 1 }\) after \(t \mathrm {~s}\) is given by \(\mathrm { v } = \mathrm { t } ^ { 3 } - 5 \mathrm { t } ^ { 2 }\).

1. Find the times at which the particle is stationary.
2. Find the total distance travelled by the particle in the first 6 seconds.

6 The displacement of a particle is modelled by the equation \(\mathrm { s } = 7 + 4 \mathrm { t } - \mathrm { t } ^ { 2 }\), where \(s\) metres is the displacement from the origin at time \(t\) seconds. The diagram shows part of the displacement-time graph for the particle. The point \(( 2,11 )\) is the maximum point on the graph. \includegraphics[max width=\textwidth, alt={}, center]{5428eabf-431d-4db1-8c25-1f2b9570d9aa-4_513_1381_422_255}

1. Kai argues that the point \(( 2,11 )\) is on the graph, so the particle has travelled a distance of 11 metres in the first 2 seconds. Comment on the validity of Kai's argument.
2. Determine the total distance the particle travels in the first 10 seconds.
3. Find an expression for the velocity of the particle at time \(t\).
4. Find the speed of the particle when \(t = 10\).

7

1. Find \(\int x ^ { 3 } \left( 15 x + \frac { 11 } { \sqrt [ 3 ] { x } } \right) \mathrm { d } x\).
2. Show that \(\int _ { 0 } ^ { 8 } x ^ { 3 } \left( 15 x + \frac { 11 } { \sqrt [ 3 ] { x } } \right) \mathrm { d } x = a \times 2 ^ { 11 }\), where \(a\) is a positive integer to be determined.

11 Given that \(k\) is a positive constant, show that \(\int _ { k } ^ { 2 k } \frac { 2 } { ( 2 x + k ) ^ { 2 } } d x\) is inversely proportional to \(k\).

8 The equation of a curve is \(\mathrm { y } = \sqrt { \sin 4 \mathrm { x } } + 2 \cos 2 \mathrm { x }\), where \(x\) is in radians.

1. Show that, for small values of \(x , y \approx 2 \sqrt { x } + 2 - 4 x ^ { 2 }\). The diagram shows the region bounded by the curve \(\mathrm { y } = \sqrt { \sin 4 \mathrm { x } } + 2 \cos 2 \mathrm { x }\), the axes and the line \(x = 0.1\). \includegraphics[max width=\textwidth, alt={}, center]{1d0ca3d5-6529-435f-a0b8-50ea4859adde-07_499_881_589_223}
2. In this question you must show detailed reasoning. Use the approximation in part (a) to estimate the area of this region.

8 Fig. 8.1 shows the cross-section of a straight driveway 4 m wide made from tarmac. \begin{figure}[h] \end{figure} The height \(h \mathrm {~m}\) of the cross-section at a displacement \(x \mathrm {~m}\) from the middle is modelled by \(\mathrm { h } = \frac { 0.2 } { 1 + \mathrm { x } ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\). A lower bound of \(0.3615 \mathrm {~m} ^ { 2 }\) is found for the area of the cross-section using rectangles as shown in Fig. 8.2. \begin{figure}[h] \end{figure}

1. Use a similar method to find an upper bound for the area of the cross-section.
2. Use the trapezium rule with 4 strips to estimate \(\int _ { 0 } ^ { 2 } \frac { 0.2 } { 1 + x ^ { 2 } } d x\).
3. The driveway is 10 m long. Use your answer in part (b) to find an estimate of the volume of tarmac needed to make the driveway.

10 In this question you must show detailed reasoning.
Show that \(\int _ { \mathrm { e } } ^ { \pi } \frac { 1 } { x } \mathrm {~d} x = \ln \pi - 1\) as given in line 37.

5 A particle \(P\) of mass 3 kg moves on the \(x\)-axis under the action of a single force acting in the positive \(x\)-direction. At time \(t \mathrm {~s}\), where \(t \geqslant 0\), the displacement of \(P\) is \(x \mathrm {~m}\) and its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The magnitude of the force acting is inversely proportional to \(( t + 1 ) ^ { 2 }\). Initially \(P\) is at rest at the point where \(x = 1\). When \(t = 1 , v = 2\).

1. Show that \(\frac { \mathrm { dv } } { \mathrm { dt } } = \frac { \mathrm { k } } { 3 ( \mathrm { t } + 1 ) ^ { 2 } }\) where \(k\) is a constant.
2. Find an expression for \(v\) in terms of \(t\).
3. Find an expression for \(x\) in terms of \(t\). As \(t\) increases, \(v\) approaches a limiting value, \(\mathrm { V } _ { \mathrm { T } }\).
4. Determine how far \(P\) is from its initial position at the instant when \(v\) is \(95 \%\) of \(\mathrm { V } _ { \mathrm { T } }\).

5 A particle \(P\) of mass \(m \mathrm {~kg}\) is projected vertically upwards through a liquid. Student \(A\) measures \(P\) 's initial speed as \(( 8.5 \pm 0.25 ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and they also record the time for \(P\) to attain its greatest height above the initial point of projection as over 3 seconds.
In an attempt to model the motion of \(P\) student \(B\) determines that \(t\) seconds after projection the only forces acting on \(P\) are its weight and the resistance from the liquid. Student \(B\) models the resistance from the liquid to be of magnitude \(m v ^ { 2 } - 6 m v\), where \(v\) is the speed of the particle.

1. (a) Show that \(\frac { \mathrm { d } t } { \mathrm {~d} v } = - \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 }\).
   (b) Determine whether student \(B\) 's model is consistent with the time recorded by \(A\) for \(P\) to attain its greatest height. After attaining its greatest height \(P\) now falls through the liquid. Student \(C\) claims that the time taken for \(P\) to achieve a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when falling through the liquid is given by $$- \int _ { 0 } ^ { 1 } \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 } \mathrm {~d} v$$
2. Explain why student \(C\) 's claim is incorrect and write down the integral which would give the correct time for \(P\) to achieve a speed of \(1 \mathrm {~ms} ^ { - 1 }\) when falling through the liquid.

5 The curve with equation \(y = x ^ { 3 } - 10 x ^ { 2 } + 28 x\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{f2c95d73-d3fe-48f7-af07-84f12bb06727-3_483_899_402_568} The curve crosses the \(x\)-axis at the origin \(O\) and the point \(A ( 3,21 )\) lies on the curve.

1. 1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence show that the curve has a stationary point when \(x = 2\) and find the \(x\)-coordinate of the other stationary point.
2. 1. Find \(\int \left( x ^ { 3 } - 10 x ^ { 2 } + 28 x \right) \mathrm { d } x\).
   2. Hence show that \(\int _ { 0 } ^ { 3 } \left( x ^ { 3 } - 10 x ^ { 2 } + 28 x \right) \mathrm { d } x = 56 \frac { 1 } { 4 }\).
   3. Hence determine the area of the shaded region bounded by the curve and the line \(O A\).

4 The curve with equation \(y = x ^ { 4 } - 8 x + 9\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-5_410_609_383_721} The point \(( 2,9 )\) lies on the curve.

1. 1. Find \(\int _ { 0 } ^ { 2 } \left( x ^ { 4 } - 8 x + 9 \right) \mathrm { d } x\).
   2. Hence find the area of the shaded region bounded by the curve and the line \(y = 9\).
2. The point \(A ( 1,2 )\) lies on the curve with equation \(y = x ^ { 4 } - 8 x + 9\).
   1. Find the gradient of the curve at the point \(A\).
   2. Hence find an equation of the tangent to the curve at the point \(A\).

6 The curve with equation \(y = x ^ { 3 } - 2 x ^ { 2 } + 3\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{c44c229e-44b2-4799-9c9c-bfccdd09d450-4_590_787_365_625} The curve cuts the \(x\)-axis at the point \(A ( - 1,0 )\) and passes through the point \(B ( 1,2 )\).

1. Find \(\int _ { - 1 } ^ { 1 } \left( x ^ { 3 } - 2 x ^ { 2 } + 3 \right) \mathrm { d } x\).
   (5 marks)
2. Hence find the area of the shaded region bounded by the curve \(y = x ^ { 3 } - 2 x ^ { 2 } + 3\) and the line \(A B\).
   (3 marks)

5

1. 1. Express \(x ^ { 2 } - 3 x + 5\) in the form \(( x - p ) ^ { 2 } + q\).
   2. Hence write down the equation of the line of symmetry of the curve with equation \(y = x ^ { 2 } - 3 x + 5\).
2. The curve \(C\) with equation \(y = x ^ { 2 } - 3 x + 5\) and the straight line \(y = x + 5\) intersect at the point \(A ( 0,5 )\) and at the point \(B\), as shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{dbc25177-4a28-480f-93d5-41acb2a2d28c-4_471_707_653_676}
   1. Find the coordinates of the point \(B\).
   2. Find \(\int \left( x ^ { 2 } - 3 x + 5 \right) \mathrm { d } x\).
   3. Find the area of the shaded region \(R\) bounded by the curve \(C\) and the line segment \(A B\).

6 A curve has equation \(y = x ^ { 5 } - 2 x ^ { 2 } + 9\). The point \(P\) with coordinates \(( - 1,6 )\) lies on the curve.

1. Find the equation of the tangent to the curve at the point \(P\), giving your answer in the form \(y = m x + c\).
2. The point \(Q\) with coordinates \(( 2 , k )\) lies on the curve.
   1. Find the value of \(k\).
   2. Verify that \(Q\) also lies on the tangent to the curve at the point \(P\).
3. The curve and the tangent to the curve at \(P\) are sketched below. \includegraphics[max width=\textwidth, alt={}, center]{aa42b4fd-1e37-48b8-90ee-269916c4db2c-4_721_887_936_589}
   1. Find \(\int _ { - 1 } ^ { 2 } \left( x ^ { 5 } - 2 x ^ { 2 } + 9 \right) \mathrm { d } x\).
   2. Hence find the area of the shaded region bounded by the curve and the tangent to the curve at \(P\).
      (3 marks)

3 The diagram shows a sketch of a curve and a line. \includegraphics[max width=\textwidth, alt={}, center]{c7f38f7e-75aa-4b41-96fd-f38f968c225c-06_520_588_351_742} The curve has equation \(y = x ^ { 4 } + 3 x ^ { 2 } + 2\). The points \(A ( - 1,6 )\) and \(B ( 2,30 )\) lie on the curve.

1. Find an equation of the tangent to the curve at the point \(A\).
2. 1. Find \(\int _ { - 1 } ^ { 2 } \left( x ^ { 4 } + 3 x ^ { 2 } + 2 \right) \mathrm { d } x\).
   2. Calculate the area of the shaded region bounded by the curve and the line \(A B\).
      [0pt] [3 marks] \(4 \quad\) A circle with centre \(C\) has equation \(x ^ { 2 } + y ^ { 2 } + 2 x - 6 y - 40 = 0\).

7 The diagram shows the sketch of a curve and the tangent to the curve at \(P\). \includegraphics[max width=\textwidth, alt={}, center]{0d5b9235-af2b-4fd5-8fcf-b2b45e3c0a3c-14_519_817_356_614} The curve has equation \(y = 4 - x ^ { 2 } - 3 x ^ { 3 }\) and the point \(P ( - 2,24 )\) lies on the curve. The tangent at \(P\) crosses the \(x\)-axis at \(Q\).

1. 1. Find the equation of the tangent to the curve at the point \(P\), giving your answer in the form \(y = m x + c\).
   2. Hence find the \(x\)-coordinate of \(Q\).
2. 1. Find \(\int _ { - 2 } ^ { 1 } \left( 4 - x ^ { 2 } - 3 x ^ { 3 } \right) \mathrm { d } x\).
   2. The point \(R ( 1,0 )\) lies on the curve. Calculate the area of the shaded region bounded by the curve and the lines \(P Q\) and \(Q R\).
      [0pt] [3 marks]

1. The curve \(C\) with equation \(y = \mathrm { f } ( x )\) is such that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 \sqrt { } x + \frac { 12 } { \sqrt { } x } , \quad x > 0$$

1. Show that, when \(x = 8\), the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is \(9 \sqrt { } 2\). The curve \(C\) passes through the point \(( 4,30 )\).
2. Using integration, find \(\mathrm { f } ( x )\).
   

1. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 5 + \frac { 1 } { x ^ { 2 } } .$$

1. Use integration to find \(y\) in terms of \(x\).
2. Given that \(y = 7\) when \(x = 1\), find the value of \(y\) at \(x = 2\).

5. The curve \(C\) with equation \(y = \mathrm { f } ( x )\) is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 \sqrt { } x + \frac { 12 } { \sqrt { } x } , x > 0\).

1. Show that, when \(x = 8\), the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is \(9 \sqrt { } 2\). The curve \(C\) passes through the point \(( 4,30 )\).
2. Using integration, find \(\mathrm { f } ( x )\).

4. 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

1. Write \(\sqrt { x }\) in the form \(x ^ { k }\), where \(k\) is a fraction.
2. Hence express \(\sqrt { x } ( x - 1 )\) in the form \(x ^ { p } - x ^ { q }\).
3. Find \(\int \sqrt { x } ( x - 1 ) \mathrm { d } x\).
4. Hence show that \(\int _ { 1 } ^ { 2 } \sqrt { x } ( x - 1 ) \mathrm { d } x = \frac { 4 } { 15 } ( \sqrt { 2 } + 1 )\).

2

1. Use the trapezium rule with five ordinates (four strips) to find an approximate value for $$\int _ { 1 } ^ { 5 } \frac { 1 } { x ^ { 2 } + 1 } \mathrm {~d} x$$ giving your answer to three significant figures.
2. 1. Find \(\int \left( x ^ { - \frac { 3 } { 2 } } + 6 x ^ { \frac { 1 } { 2 } } \right) \mathrm { d } x\), giving the coefficient of each term in its simplest form.
   2. Hence find the value of \(\int _ { 1 } ^ { 4 } \left( x ^ { - \frac { 3 } { 2 } } + 6 x ^ { \frac { 1 } { 2 } } \right) \mathrm { d } x\).