1.08e Area between curve and x-axis: using definite integrals

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 \(\frac { \text { Area to left of curve } } { \text { Area below curve } } = 5\).

8 In this question you must show detailed reasoning. Fig. 8 shows the graph of a quadratic function. The graph crosses the axes at the points \(( - 1,0 ) , ( 0 , - 4 )\) and \(( 2,0 )\). \begin{figure}[h] \end{figure} Find the area of the finite region bounded by the curve and the \(x\)-axis.

12 In this question you must show detailed reasoning. Fig. 12 shows part of the graph of \(y = x ^ { 2 } + \frac { 1 } { x ^ { 2 } }\). \begin{figure}[h] \end{figure} The tangent to the curve \(\mathrm { y } = \mathrm { x } ^ { 2 } + \frac { 1 } { \mathrm { x } ^ { 2 } }\) at the point \(\left( 2 , \frac { 17 } { 4 } \right)\) meets the \(x\)-axis at A and meets the \(y\)-axis at B . O is the origin.

1. Find the exact area of the triangle OAB .
2. Use calculus to prove that the complete curve has two minimum points and no maximum point. \section*{END OF QUESTION PAPER}

10 In this question you must show detailed reasoning. The equation of a curve is \(y = \frac { x ^ { 2 } } { 4 } + \frac { 2 } { x } + 1\). A tangent and a normal to the curve are drawn at the point where \(x = 2\). Calculate the area bounded by the tangent, the normal and the \(x\)-axis. \section*{END OF QUESTION PAPER}

5 The curve \(C\) has equation $$3 x ^ { 2 } - 5 x y + \mathrm { e } ^ { 2 y - 4 } + 6 = 0$$ The point \(P\) with coordinates \(( 1,2 )\) lies on \(C\). The tangent to \(C\) at \(P\) meets the \(y\)-axis at the point \(A\) and the normal to \(C\) at \(P\) meets the \(y\)-axis at the point \(B\). Find the exact area of triangle \(A B P\).

5

1. Find \(\int \left( x ^ { 3 } - 6 x \right) \mathrm { d } x\).
2. 1. Find \(\int \left( \frac { 4 } { x ^ { 2 } } - 1 \right) \mathrm { d } x\).
   2. The diagram shows part of the curve \(y = \frac { 4 } { x ^ { 2 } } - 1\). \includegraphics[max width=\textwidth, alt={}, center]{35d8bb6d-ff0f-4590-b13d-46e4869e2587-04_707_1283_708_415} The curve crosses the \(x\)-axis at \(( 2,0 )\).
      The shaded region is bounded by the curve, the \(x\)-axis, and the lines \(x = 1\) and \(x = 5\). Calculate the area of the shaded region.

6 The curve with equation \(y = 3 x ^ { 5 } + 2 x + 5\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{33da89e2-f74f-4d5a-8bbd-ceaa728b6c34-5_428_563_372_740} The curve cuts the \(x\)-axis at the point \(A ( - 1,0 )\) and cuts the \(y\)-axis at the point \(B\).

1. 1. State the coordinates of the point \(B\) and hence find the area of the triangle \(A O B\), where \(O\) is the origin.
   2. Find \(\int \left( 3 x ^ { 5 } + 2 x + 5 \right) \mathrm { d } x\).
   3. Hence find the area of the shaded region bounded by the curve and the line \(A B\).
2. 1. Find the gradient of the curve with equation \(y = 3 x ^ { 5 } + 2 x + 5\) at the point \(A ( - 1,0 )\).
   2. Hence find an equation of the tangent to the curve at the point \(A\).

6

1. The polynomial \(\mathrm { p } ( x )\) is given by \(\mathrm { p } ( x ) = x ^ { 3 } - 7 x - 6\).
   1. Use the Factor Theorem to show that \(x + 1\) is a factor of \(\mathrm { p } ( x )\).
   2. Express \(\mathrm { p } ( x ) = x ^ { 3 } - 7 x - 6\) as the product of three linear factors.
2. The curve with equation \(y = x ^ { 3 } - 7 x - 6\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{de4f827d-f237-488a-9177-3d85d0cb1771-4_403_762_651_641} The curve cuts the \(x\)-axis at the point \(A\) and the points \(B ( - 1,0 )\) and \(C ( 3,0 )\).
   1. State the coordinates of the point \(A\).
   2. Find \(\int _ { - 1 } ^ { 3 } \left( x ^ { 3 } - 7 x - 6 \right) \mathrm { d } x\).
   3. Hence find the area of the shaded region bounded by the curve \(y = x ^ { 3 } - 7 x - 6\) and the \(x\)-axis between \(B\) and \(C\).
   4. Find the gradient of the curve \(y = x ^ { 3 } - 7 x - 6\) at the point \(B\).
   5. Hence find an equation of the normal to the curve at the point \(B\).

6

1. The polynomial \(\mathrm { f } ( x )\) is given by \(\mathrm { f } ( x ) = x ^ { 3 } + 4 x - 5\).
   1. Use the Factor Theorem to show that \(x - 1\) is a factor of \(\mathrm { f } ( x )\).
   2. Express \(\mathrm { f } ( x )\) in the form \(( x - 1 ) \left( x ^ { 2 } + p x + q \right)\), where \(p\) and \(q\) are integers.
   3. Hence show that the equation \(\mathrm { f } ( x ) = 0\) has exactly one real root and state its value.
2. The curve with equation \(y = x ^ { 3 } + 4 x - 5\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{23f34515-3373-4644-a8a1-82b45809d934-4_505_959_868_529} The curve cuts the \(x\)-axis at the point \(A ( 1,0 )\) and the point \(B ( 2,11 )\) lies on the curve.
   1. Find \(\int \left( x ^ { 3 } + 4 x - 5 \right) \mathrm { d } x\).
   2. Hence find the area of the shaded region bounded by the curve and the line \(A B\).

5 The curve with equation \(y = 16 - x ^ { 4 }\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{fddf5016-a5bd-42db-b5c4-f4980b8d9d67-3_435_663_824_685} The points \(A ( - 2,0 ) , B ( 2,0 )\) and \(C ( 1,15 )\) lie on the curve.

1. Find an equation of the straight line \(A C\).
2. 1. Find \(\int _ { - 2 } ^ { 1 } \left( 16 - x ^ { 4 } \right) \mathrm { d } x\).
   2. Hence calculate the area of the shaded region bounded by the curve and the line \(A C\).

4

1. The polynomial \(\mathrm { p } ( x )\) is given by \(\mathrm { p } ( x ) = x ^ { 3 } - x + 6\).
   1. Find the remainder when \(\mathrm { p } ( x )\) is divided by \(x - 3\).
   2. Use the Factor Theorem to show that \(x + 2\) is a factor of \(\mathrm { p } ( x )\).
   3. Express \(\mathrm { p } ( x ) = x ^ { 3 } - x + 6\) in the form \(( x + 2 ) \left( x ^ { 2 } + b x + c \right)\), where \(b\) and \(c\) are integers.
   4. The equation \(\mathrm { p } ( x ) = 0\) has one root equal to - 2 . Show that the equation has no other real roots.
2. The curve with equation \(y = x ^ { 3 } - x + 6\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{5f1ff5fa-b6e8-4c4f-aef7-63eb947b299f-3_529_702_945_667} The curve cuts the \(x\)-axis at the point \(A ( - 2,0 )\) and the \(y\)-axis at the point \(B\).
   1. State the \(y\)-coordinate of the point \(B\).
   2. Find \(\int _ { - 2 } ^ { 0 } \left( x ^ { 3 } - x + 6 \right) \mathrm { d } x\).
   3. Hence find the area of the shaded region bounded by the curve \(y = x ^ { 3 } - x + 6\) and the line \(A B\).

6 A curve \(C\) is defined for \(x > 0\) by the equation \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\) and is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-4_545_784_420_628}

1. 1. Given that \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. The curve \(C\) has a minimum point \(M\). Find the coordinates of \(M\).
   3. Find an equation of the normal to \(C\) at the point ( 1,6 ).
2. 1. Find \(\int \left( x + 1 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x\).
   2. Hence find the area of the region bounded by the curve \(C\), the lines \(x = 1\) and \(x = 4\) and the \(x\)-axis.

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 \(4 \frac { 1 } { 2 }\).

9

1. Given that \(y = x ^ { - 2 } \ln x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 \ln x } { x ^ { 3 } }\).
2. Using integration by parts, find \(\int x ^ { - 2 } \ln x \mathrm {~d} x\).
3. The sketch shows the graph of \(y = x ^ { - 2 } \ln x\). \includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-06_604_1045_687_536}
   1. Using the answer to part (a), find, in terms of e, the \(x\)-coordinate of the stationary point \(A\).
   2. The region \(R\) is bounded by the curve, the \(x\)-axis and the line \(x = 5\). Using your answer to part (b), show that the area of \(R\) is $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) . The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) . Angle \(B A E =\) angle \(B C D = \theta\) radians. The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed,with the added restriction that none of the line segments cross.

1. Sketch \(S ( \pi )\) ,labelling the vertices clearly. The shape \(S ( \phi )\) is a trapezium.
2. Sketch \(S ( \phi )\) and calculate the value of \(\phi\) . The smallest possible value for \(\theta\) is \(\alpha\) ,where \(\alpha > 0\) ,and the largest possible value for \(\theta\) is \(\beta\) , where \(\beta > \pi\) .
3. Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
4. Find the value of \(\beta\) . The area,in \(\mathrm { cm } ^ { 2 }\) ,of shape \(S ( \theta )\) is \(R ( \theta )\) .
5. Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
6. show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
7. find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
   END

3

1. Amaya and Ben integrated \(( 1 + x ) ^ { 2 }\), with respect to \(x\), using different methods, as follows. Amaya: \(\quad \int ( 1 + x ) ^ { 2 } \mathrm {~d} x = \frac { ( 1 + x ) ^ { 3 } } { 3 } + c \quad = \frac { 1 } { 3 } + x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + c\) Ben: \(\quad \int ( 1 + x ) ^ { 2 } \mathrm {~d} x = \int \left( 1 + 2 x + x ^ { 2 } \right) \mathrm { d } x = x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + c\) Charlie said that, because these answers are different, at least one of them must be wrong. Explain whether you agree with Charlie's statement.
2. You are given that \(a\) is a constant greater than 1 .
   1. Find \(\int _ { 1 } ^ { a } \frac { 1 } { ( 1 + x ) ^ { 2 } } \mathrm {~d} x\), giving your answer as a single fraction in terms of the constant \(a\).
   2. You are given that the area enclosed by the curve \(y = \frac { 1 } { ( 1 + x ) ^ { 2 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = a\) is equal to \(\frac { 1 } { 3 }\). Determine the value of \(a\).
3. In this question you must show detailed reasoning. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 12 } \pi } \frac { \cos 2 x } { \sin 2 x + 2 } \mathrm {~d} x\), giving your answer in its simplest form.

6 A curve has equation \(y = \frac { 2 } { x \sqrt { x } }\) \includegraphics[max width=\textwidth, alt={}, center]{b45dc98e-1699-47c9-9228-5abe0e5c9195-05_508_549_420_744} The region enclosed between the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = a\) has area 3 units. Given that \(a > 1\), find the value of \(a\).
Fully justify your answer.

8 A curve has equation $$y = x ^ { 3 } + p x ^ { 2 } + q x - 45$$ The curve passes through point \(R ( 2,3 )\) The gradient of the curve at \(R\) is 8
8

1. Find the value of \(p\) and the value of \(q\).
   8
2. Calculate the area enclosed between the normal to the curve at \(R\) and the coordinate 8 (b) axes. \(9 \quad\) A curve \(C\) has equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = ( x - 2 ) ( x - 3 ) ^ { 2 }$$

7 The diagram below shows the graph of the curve that has equation \(y = x ^ { 2 } - 3 x + 2\) along with two shaded regions. \includegraphics[max width=\textwidth, alt={}, center]{f87d1b36-26db-4a0b-b9ec-d7d82a396aba-08_646_711_408_667} 7

1. State the coordinates of the points \(A , B\) and \(C\).
   7
2. Katy is asked by her teacher to find the total area of the two shaded regions.
   Katy uses her calculator to find \(\int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 3 x + 2 \right) \mathrm { d } x\) and gets the answer \(\frac { 2 } { 3 }\) Katy's teacher says that her answer is incorrect.
   7 (b) (i) Show that the total area of the two shaded regions is 1
   Fully justify your answer.
   7 (b) (ii) Explain why Katy's method was not valid.

9 The diagram below shows the graphs of \(y = x ^ { 2 } - 4 x - 12\) and \(y = x + 2\) \includegraphics[max width=\textwidth, alt={}, center]{11168e8f-5ba5-4d27-83ab-0327cc23d08c-10_933_912_358_566} 9

1. Write down three inequalities which together describe the shaded region.
   9
2. Find the coordinates of the points \(A , B\) and \(C\).
   9
3. Find the exact area of the shaded region.
   Fully justify your answer.
   [0pt] [6 marks]

11 The region \(R\) enclosed by the lines \(x = 1 , x = 6 , y = 0\) and the curve $$y = \ln ( 8 - x )$$ is shown shaded in Figure 3 below. \begin{figure}[h] \end{figure} All distances are measured in centimetres.
11

1. Use a single trapezium to find an approximate value of the area of the shaded region, giving your answer in \(\mathrm { cm } ^ { 2 }\) to two decimal places.
   [0pt] [2 marks]
   \section*{Question 11 continues on the next page} 11
2. Shape \(B\) is made from four copies of region \(R\) as shown in Figure 4 below. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-18_707_711_438_667}
   \end{figure} Shape \(B\) is cut from metal of thickness 2 mm
   The metal has a density of \(10.5 \mathrm {~g} / \mathrm { cm } ^ { 3 }\) Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B. Give your answer to the nearest gram.
   11
3. Without further calculation, give one reason why the mass found in part (b) may be:
   11 (c) (i) an underestimate.
   11 (c) (ii) an overestimate.

15 The region enclosed between the curves \(y = \mathrm { e } ^ { x } , y = 6 - \mathrm { e } ^ { \overline { 2 } }\) and the line \(x = 0\) is shown shaded in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-28_1155_1009_424_513} Show that the exact area of the shaded region is $$6 \ln 4 - 5$$ Fully justify your answer. \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-30_2491_1736_219_139}

10

1. Given that $$y = \tan x$$ use the quotient rule to show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec ^ { 2 } x$$ 10
2. The region enclosed by the curve \(y = \tan ^ { 2 } x\) and the horizontal line, which intersects the curve at \(x = - \frac { \pi } { 4 }\) and \(x = \frac { \pi } { 4 }\), is shaded in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-17_1059_967_461_539} Show that the area of the shaded region is $$\pi - 2$$ Fully justify your answer.
   Do not write outside the box \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-19_2488_1716_219_153}

15

1. Show that $$\sin x - \sin x \cos 2 x \approx 2 x ^ { 3 }$$ for small values of \(x\).
   15
2. Hence, show that the area between the graph with equation $$y = \sqrt { 8 ( \sin x - \sin x \cos 2 x ) }$$ the positive \(x\)-axis and the line \(x = 0.25\) can be approximated by $$\text { Area } \approx 2 ^ { m } \times 5 ^ { n }$$ where \(m\) and \(n\) are integers to be found.
   15
3. (i) Explain why $$\int _ { 6.3 } ^ { 6.4 } 2 x ^ { 3 } \mathrm {~d} x$$ is not a suitable approximation for $$\int _ { 6.3 } ^ { 6.4 } ( \sin x - \sin x \cos 2 x ) d x$$ Question 15 continues on the next page 15 (c) (ii) Explain how $$\int _ { 6.3 } ^ { 6.4 } ( \sin x - \sin x \cos 2 x ) d x$$ may be approximated by $$\int _ { a } ^ { b } 2 x ^ { 3 } \mathrm {~d} x$$ for suitable values of \(a\) and \(b\). \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-31_2492_1721_217_150}
   \includegraphics[max width=\textwidth, alt={}]{042e248a-9efa-4844-957d-f05715900ffc-36_2486_1719_221_150}

7 The curve \(y = 15 - x ^ { 2 }\) and the isosceles triangle \(O P Q\) are shown on the diagram The curve \(y = 15 - x ^ { 2 }\) and the isosceles triangle \(O P Q\) are shown on the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-10_759_810_388_614} Vertices \(P\) and \(Q\) lie on the curve such that \(Q\) lies vertically above some point ( \(q , 0\) ) The line \(P Q\) is parallel to the \(x\)-axis. 7

1. Show that the area, \(A\), of the triangle \(O P Q\) is given by $$A = 15 q - q ^ { 3 } \quad \text { for } 0 < q < c$$ where \(c\) is a constant to be found.
   7
2. Find the exact maximum area of triangle \(O P Q\). Fully justify your answer.