1.08f Area between two curves: using integration

7

1. Find \(\int \left( x ^ { 2 } + 4 \right) ( x - 6 ) \mathrm { d } x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{ad3083ae-caa6-42d8-a1f2-e984150cb104-4_449_551_349_758} The diagram shows the curve \(y = 6 x ^ { \frac { 3 } { 2 } }\) and part of the curve \(y = \frac { 8 } { x ^ { 2 } } - 2\), which intersect at the point \(( 1,6 )\). Use integration to find the area of the shaded region enclosed by the two curves and the \(x\)-axis.

4 \includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-3_588_1136_255_502} The diagram shows the curve \(y = - 1 + \sqrt { x + 4 }\) and the line \(y = 3\).

1. Show that \(y = - 1 + \sqrt { x + 4 }\) can be rearranged as \(x = y ^ { 2 } + 2 y - 3\).
2. Hence find by integration the exact area of the shaded region enclosed between the curve, the \(y\)-axis and the line \(y = 3\).

8 \includegraphics[max width=\textwidth, alt={}, center]{c940af95-e291-402a-856c-9090d13163d5-4_538_702_264_719} The diagram shows the curve with equation $$y = \frac { 6 } { \sqrt { x } } - 3$$ The point \(P\) has coordinates \(( 0 , p )\). The shaded region is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = p\). The shaded region is rotated completely about the \(y\)-axis to form a solid of volume \(V\).

1. Show that \(V = 16 \pi \left( 1 - \frac { 27 } { ( p + 3 ) ^ { 3 } } \right)\).
2. It is given that \(P\) is moving along the \(y\)-axis in such a way that, at time \(t\), the variables \(p\) and \(t\) are related by $$\frac { \mathrm { d } p } { \mathrm {~d} t } = \frac { 1 } { 3 } p + 1 .$$ Find the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\) at the instant when \(p = 9\).

9 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-4_661_915_269_557} The diagram shows the curve $$y = \mathrm { e } ^ { 2 x } - 18 x + 15 .$$ The curve crosses the \(y\)-axis at \(P\) and the minimum point is \(Q\). The shaded region is bounded by the curve and the line \(P Q\).

1. Show that the \(x\)-coordinate of \(Q\) is \(\ln 3\).
2. Find the exact area of the shaded region.

5 \includegraphics[max width=\textwidth, alt={}, center]{6d15cb4d-f540-488b-b94e-7a494f192ba5-2_469_721_1932_662} The diagram shows the curves \(y = \mathrm { e } ^ { 2 x }\) and \(y = 8 \mathrm { e } ^ { - x }\). The shaded region is bounded by the curves and the \(y\)-axis. Without using a calculator,

1. solve an appropriate equation to show that the curves intersect at a point for which \(x = \ln 2\),
2. find the area of the shaded region, giving your answer in simplified form.

10

1. Express \(\frac { x + 8 } { x ( x + 2 ) }\) in partial fractions.
2. By first using division, express \(\frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\) in the form \(P + \frac { Q } { x } + \frac { R } { x + 2 }\). A curve has parametric equations \(x = \frac { 2 t } { 1 - t } , y = 3 t + \frac { 4 } { t }\).
3. Show that the cartesian equation of the curve is \(y = \frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\).
4. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\). Give your answer in the form \(L + M \ln 2 + N \ln 3\).

3 A finite region \(R\) in the \(x - y\) plane is bounded by the curve with equation \(y = \sqrt { } x - \frac { 1 } { \sqrt { } x }\), the \(x\)-axis between \(x = 1\) and \(x = 4\), and the line \(x = 4\). Find the exact value of the \(y\)-coordinate of the centroid of \(R\).

12 In this question you must show detailed reasoning. \includegraphics[max width=\textwidth, alt={}, center]{1ba9fa5f-310f-4429-9bd1-4004852d5b3e-6_716_479_292_794} The diagram shows the curve \(y = \frac { 4 \cos 2 x } { 3 - \sin 2 x }\), for \(x \geqslant 0\), and the normal to the curve at the point \(\left( \frac { 1 } { 4 } \pi , 0 \right)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac { 9 } { 4 } + \frac { 1 } { 128 } \pi ^ { 2 }\).
[0pt] [10]

11
The diagram shows part of the curve \(y = \ln ( x - 4 )\).

1. Use integration by parts to show that \(\int \ln ( x - 4 ) \mathrm { d } x = ( x - 4 ) \ln | x - 4 | - x + c\).
2. State the equation of the vertical asymptote to the curve \(y = \ln ( x - 4 )\).
3. Find the total area enclosed by the curve \(y = \ln ( x - 4 )\), the \(x\)-axis and the lines \(x = 4.5\) and \(x = 7\). Give your answer in the form \(a \ln 3 + b \ln 2 + c\) where \(a , b\) and \(c\) are constants to be found.

5 The diagram shows the circle with centre O and radius 2, and the parabola \(y = \frac { 1 } { \sqrt { 3 } } \left( 4 - x ^ { 2 } \right)\). \includegraphics[max width=\textwidth, alt={}, center]{f2f45d6c-cfdc-455b-ab08-597b06a69f36-06_838_970_1059_280} The circle meets the parabola at points \(P\) and \(Q\), as shown in the diagram.

1. Verify that the coordinates of \(Q\) are \(( 1 , \sqrt { 3 } )\).
2. Find the exact area of the shaded region enclosed by the \(\operatorname { arc } P Q\) of the circle and the parabola.
   [0pt] [8]

12. \begin{figure}[h] \end{figure} Figure 5 shows the plan view of the design for a swimming pool.
The pool is modelled as a quarter of a circle joined to two equal sized rectangles as shown. Given that

• the quarter circle has radius \(x\) metres
• the rectangles each have length \(x\) metres and width \(y\) metres
• the total surface area of the swimming pool is \(100 \mathrm {~m} ^ { 2 }\)
  1. show that, according to the model, the perimeter \(P\) metres of the swimming pool is given by

$$P = 2 x + \frac { 200 } { x }$$

Use calculus to find the value of \(x\) for which \(P\) has a stationary value.

Prove, by further calculus, that this value of \(x\) gives a minimum value for \(P\) Access to the pool is by side \(A B\) shown in Figure 5.
Given that \(A B\) must be at least one metre,

determine, according to the model, whether the swimming pool with the minimum perimeter would be suitable.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = x ( x + 2 ) ( x - 4 )\).
The region \(R _ { 1 }\) shown shaded in Figure 2 is bounded by the curve and the negative \(x\)-axis.

1. Show that the exact area of \(R _ { 1 }\) is \(\frac { 20 } { 3 }\) The region \(R _ { 2 }\) also shown shaded in Figure 2 is bounded by the curve, the positive \(x\)-axis and the line with equation \(x = b\), where \(b\) is a positive constant and \(0 < b < 4\) Given that the area of \(R _ { 1 }\) is equal to the area of \(R _ { 2 }\)
2. verify that \(b\) satisfies the equation $$( b + 2 ) ^ { 2 } \left( 3 b ^ { 2 } - 20 b + 20 \right) = 0$$ The roots of the equation \(3 b ^ { 2 } - 20 b + 20 = 0\) are 1.225 and 5.442 to 3 decimal places. The value of \(b\) is therefore 1.225 to 3 decimal places.
3. Explain, with the aid of a diagram, the significance of the root 5.442

1. In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of a curve with equation $$y = \frac { ( x - 2 ) ( x - 4 ) } { 4 \sqrt { x } } \quad x > 0$$ The region \(R\), shown shaded in Figure 3, is bounded by the curve and the \(x\)-axis.
Find the exact area of \(R\), writing your answer in the form \(a \sqrt { 2 } + b\), where \(a\) and \(b\) are constants to be found.

6 \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-05_538_531_264_246} The shape \(A B C\) shown in the diagram is a student's design for the sail of a small boat.
The curve \(A C\) has equation \(y = 2 \log _ { 2 } x\) and the curve \(B C\) has equation \(y = \log _ { 2 } \left( x - \frac { 3 } { 2 } \right) + 3\).

1. State the \(x\)-coordinate of point \(A\).
2. Determine the \(x\)-coordinate of point \(B\).
3. By solving an equation involving logarithms, show that the \(x\)-coordinate of point \(C\) is 2 . It is given that, correct to 3 significant figures, the area of the sail is 0.656 units \(^ { 2 }\).
4. Calculate by how much the area is over-estimated or under-estimated when the curved edges of the sail are modelled as straight lines.

11 A sports car accelerates along a straight road from rest. After 5 s its velocity is \(9 \mathrm {~ms} ^ { - 1 }\). In model A, the acceleration is assumed to be constant.

1. Calculate the distance travelled by the car in the first 5 seconds according to model A . In model B , the velocity \(v\) in \(\mathrm { ms } ^ { - 1 }\) is given by \(\mathrm { v } = 0.05 \mathrm { t } ^ { 3 } + \mathrm { kt }\), where \(t\) is the time in seconds after the start and \(k\) is a constant.
2. Find the value of \(k\) which gives the correct value of \(v\) when \(t = 5\).
3. Using this value of \(k\) in model B , calculate the acceleration of the car when \(t = 5\). The car travels 16 m in the first 5 seconds.
4. Show that model B, with the value of \(k\) found in part (b), better fits this information than model A does.

10 In this question you must show detailed reasoning.
The equation of a curve is \(y = 12 x ^ { 3 } - 24 x ^ { 2 } - 60 x + 72\).
Determine the magnitude of the total area bounded by the curve and the \(x\)-axis.

9 The diagram shows the curve \(\mathrm { y } = 3 - \sqrt { \mathrm { x } }\). \includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-6_810_1008_1155_283}

1. Draw the line \(\mathrm { y } = 5 \mathrm { x } - 1\) on the copy of the diagram in the Printed Answer Booklet.
2. In this question you must show detailed reasoning. Determine the exact area of the region bounded by the curve \(y = 3 - \sqrt { x }\), the lines \(y = 5 x - 1\) and \(x = 4\) and the \(x\)-axis.

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 \(O A\).

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)

9 The diagram shows part of a curve crossing the \(x\)-axis at the origin \(O\) and at the point \(A ( 8,0 )\). Tangents to the curve at \(O\) and \(A\) meet at the point \(P\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{02e5dfac-18d7-480d-ac23-dfd2ca348cba-5_547_536_497_760} The curve has equation $$y = 12 x - 3 x ^ { \frac { 5 } { 3 } }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. 1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(O\) and hence write down an equation of the tangent at \(O\).
   2. Show that the equation of the tangent at \(A ( 8,0 )\) is \(y + 8 x = 64\).
3. Find \(\int \left( 12 x - 3 x ^ { \frac { 5 } { 3 } } \right) \mathrm { d } x\).
4. Calculate the area of the shaded region bounded by the curve from \(O\) to \(A\) and the tangents \(O P\) and \(A P\).

7. A geometric series is \(a + a r + a r ^ { 2 } + \ldots\)

1. Prove that the sum of the first \(n\) terms of this series is given by \(S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }\). The second and fourth terms of the series are 3 and 1.08 respectively.
   Given that all terms in the series are positive, find
2. the value of \(r\) and the value of \(a\),
3. the sum to infinity of the series. \includegraphics[max width=\textwidth, alt={}, center]{1033051d-18bf-4734-a556-4c8e1c789992-4_764_1159_294_299} Fig. 2 shows part of the curve with equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 9 x\). The curve touches the \(x\)-axis at \(A\) and has a maximum turning point at \(B\).
   1. Show that the equation of the curve may be written as \(y = x ( x - 3 ) ^ { 2 }\), and hence write down the coordinates of \(A\).
   2. Find the coordinates of \(B\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.
   3. Find the area of \(R\).

7. \begin{figure}[h] \end{figure} Fig. 2 shows part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 5 x\).
The curve crosses the \(x\)-axis at the origin \(O\) and at the points \(A\) and \(B\).

1. Factorise \(\mathrm { f } ( x )\) completely.
2. Write down the \(x\)-coordinates of the points \(A\) and \(B\).
3. Find the gradient of \(C\) at \(A\). The region \(R\) is bounded by \(C\) and the line \(O A\), and the region \(S\) is bounded by \(C\) and the line \(A B\).
4. Use integration to find the area of the combined regions \(R\) and \(S\), shown shaded in Fig.2. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{e4d38009-aa70-4c55-9765-c54044ffaa31-4_736_727_338_402}
   \end{figure} Fig. 3 shows a sketch of part of the curve \(C\) with equation \(y = x ^ { 3 } - 7 x ^ { 2 } + 15 x + 3 , x \geq 0\). The point \(P\), on \(C\), has \(x\)-coordinate 1 and the point \(Q\) is the minimum turning point of \(C\).
   1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Find the coordinates of \(Q\).
   3. Show that \(P Q\) is parallel to the \(x\)-axis.
   4. Calculate the area, shown shaded in Fig. 3, bounded by \(C\) and the line \(P Q\).

8. \begin{figure}[h] \end{figure} A table top, in the shape of a parallelogram, is made from two types of wood. The design is shown in Fig. 1. The area inside the ellipse is made from one type of wood, and the surrounding area is made from a second type of wood. The ellipse has parametric equations, $$x = 5 \cos \theta , \quad y = 4 \sin \theta , \quad 0 \leq \theta < 2 \pi$$ The parallelogram consists of four line segments, which are tangents to the ellipse at the points where \(\theta = \alpha , \theta = - \alpha , \theta = \pi - \alpha , \theta = - \pi + \alpha\).

1. Find an equation of the tangent to the ellipse at ( \(5 \cos \alpha , 4 \sin \alpha\) ), and show that it can be written in the form $$5 y \sin \alpha + 4 x \cos \alpha = 20 .$$
2. Find by integration the area enclosed by the ellipse.
3. Hence show that the area enclosed between the ellipse and the parallelogram is $$\frac { 80 } { \sin 2 \alpha } - 20 \pi$$
4. Given that \(0 < \alpha < \frac { \pi } { 4 }\), find the value of \(\alpha\) for which the areas of two types of wood are equal.