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

1. In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation $$y = \frac { 1 } { 2 } x ^ { 2 } + \frac { 1458 } { \sqrt { x ^ { 3 } } } - 74 \quad x > 0$$ The point \(P\) is the only stationary point on the curve.

1. Use calculus to show that the \(x\) coordinate of \(P\) is 9 The line \(l\) passes through the point \(P\) and is parallel to the \(x\)-axis.
   The region \(R\), shown shaded in Figure 2, is bounded by the curve, the line \(l\) and the line with equation \(x = 4\)
2. Use algebraic integration to find the exact area of \(R\).

10. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { 36 } { x ^ { 2 } } + 2 x - 13 \quad x > 0$$ Using calculus,

1. find the range of values of \(x\) for which \(\mathrm { f } ( x )\) is increasing,
2. show that \(\int _ { 2 } ^ { 9 } \left( \frac { 36 } { x ^ { 2 } } + 2 x - 13 \right) \mathrm { d } x = 0\) The point \(P ( 2,0 )\) and the point \(Q ( 6,0 )\) lie on \(C\).
   Given \(\int _ { 2 } ^ { 6 } \left( \frac { 36 } { x ^ { 2 } } + 2 x - 13 \right) \mathrm { d } x = - 8\)
3. 1. state the value of \(\int _ { 6 } ^ { 9 } \left( \frac { 36 } { x ^ { 2 } } + 2 x - 13 \right) \mathrm { d } x\)
   2. find the value of the constant \(k\) such that \(\int _ { 2 } ^ { 6 } \left( \frac { 36 } { x ^ { 2 } } + 2 x + k \right) \mathrm { d } x = 0\)

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the graph of the curves \(C _ { 1 }\) and \(C _ { 2 }\) The curves intersect when \(x = 2.5\) and when \(x = 4\) A table of values for some points on the curve \(C _ { 1 }\) is shown below, with \(y\) values given to 3 decimal places as appropriate. Using the trapezium rule with all the values of \(y\) in the table,

1. find, to 2 decimal places, an estimate for the area bounded by the curve \(C _ { 1 }\), the line with equation \(x = 2.5\), the \(x\)-axis and the line with equation \(x = 4\) The curve \(C _ { 2 }\) has equation $$y = x ^ { \frac { 3 } { 2 } } - 3 x + 9 \quad x > 0$$
2. Find \(\int \left( x ^ { \frac { 3 } { 2 } } - 3 x + 9 \right) \mathrm { d } x\) The region \(R\), shown shaded in Figure 2, is bounded by the curves \(C _ { 1 }\) and \(C _ { 2 }\)
3. Use the answers to part (a) and part (b) to find, to one decimal place, an estimate for the area of the region \(R\).
   (3)

8. Solutions relying on calculator technology are not acceptable in this question.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bfeb1724-9a00-4a36-9606-520395792b2b-22_556_822_351_561} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of part of a curve with equation $$y = \frac { 8 \sqrt { x } - 5 } { 2 x ^ { 2 } } \quad x > 0$$ The region \(R\), shown shaded in Figure 2, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 4\) Find the exact area of \(R\).
2. Find the value of the constant \(k\) such that $$\int _ { - 3 } ^ { 6 } \left( \frac { 1 } { 2 } x ^ { 2 } + k \right) \mathrm { d } x = 55$$

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curves \(C _ { 1 }\) and \(C _ { 2 }\) with equations $$\begin{array} { l l } C _ { 1 } : y = x ^ { 3 } - 6 x + 9 & x \geqslant 0 \\ C _ { 2 } : y = - 2 x ^ { 2 } + 7 x - 1 & x \geqslant 0 \end{array}$$ The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the points \(A\) and \(B\) as shown in Figure 1 .
The point \(A\) has coordinates (1,4). Using algebra and showing all steps of your working,

1. find the coordinates of the point \(B\). The finite region \(R\), shown shaded in Figure 1, is bounded by \(C _ { 1 }\) and \(C _ { 2 }\)
2. Use algebraic integration to find the exact area of \(R\).

8. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 4 } { 3 } x ^ { 3 } - 11 x ^ { 2 } + k x \quad \text { where } k \text { is a constant }$$ The point \(M\) is the maximum turning point of \(C\) and is shown in Figure 2.
Given that the \(x\) coordinate of \(M\) is 2

1. show that \(k = 28\)
2. Determine the range of values of \(x\) for which \(y\) is increasing. The line \(l\) passes through \(M\) and is parallel to the \(x\)-axis.
   The region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\), the line \(l\) and the \(y\)-axis.
3. Find, by algebraic integration, the exact area of \(R\).

1. The curve \(C _ { 1 }\) has equation \(y = \mathrm { f } ( x )\).

A table of values of \(x\) and \(y\) for \(y = \mathrm { f } ( x )\) is shown below, with the \(y\) values rounded to 4 decimal places where appropriate.

1. Use the trapezium rule with all the values of \(y\) in the table to find an approximation for $$\int _ { 0 } ^ { 2 } f ( x ) d x$$ giving your answer to 3 decimal places. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6f926d53-c6de-4eb7-9d18-596f61ec26e1-16_629_592_1105_402} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6f926d53-c6de-4eb7-9d18-596f61ec26e1-16_540_456_1194_1192} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by
   • the curve \(C _ { 1 }\)
   • the curve \(C _ { 2 }\) with equation \(y = 2 - \frac { 1 } { 4 } x ^ { 2 }\)
   • the line with equation \(x = 2\)
   • the \(y\)-axis
   The region \(R\) forms part of the design for a logo shown in Figure 2.
   The design consists of the shaded region \(R\) inside a rectangle of width 2 and height 3 Using calculus and the answer to part (a),
2. calculate an estimate for the percentage of the logo which is shaded.

1. The curve \(C\) has equation

$$y = \frac { 12 x ^ { 3 } ( x - 7 ) + 14 x ( 13 x - 15 ) } { 21 \sqrt { x } } \quad x > 0$$

1. Write the equation of \(C\) in the form $$y = a x ^ { \frac { 7 } { 2 } } + b x ^ { \frac { 5 } { 2 } } + c x ^ { \frac { 3 } { 2 } } + d x ^ { \frac { 1 } { 2 } }$$ where \(a , b , c\) and \(d\) are fully simplified constants. The curve \(C\) has three turning points.
   Using calculus,
2. show that the \(x\) coordinates of the three turning points satisfy the equation $$2 x ^ { 3 } - 10 x ^ { 2 } + 13 x - 5 = 0$$ Given that the \(x\) coordinate of one of the turning points is 1
3. find, using algebra, the exact \(x\) coordinates of the other two turning points.
   (Solutions based entirely on calculator technology are not acceptable.)

1. In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 3 shows

• the curve \(C _ { 1 }\) with equation \(y = x ^ { 3 } - 5 x ^ { 2 } + 3 x + 14\)
• the circle \(C _ { 2 }\) with centre \(T\)

The point \(T\) is the minimum turning point of \(C _ { 1 }\) Using Figure 3 and calculus,

1. find the coordinates of \(T\) The curve \(C _ { 1 }\) intersects the circle \(C _ { 2 }\) at the point \(A\) with \(x\) coordinate 2
2. Find an equation of the circle \(C _ { 2 }\) The line \(l\) shown in Figure 3, is the tangent to circle \(C _ { 2 }\) at \(A\)
3. Show that an equation of \(l\) is $$y = \frac { 1 } { 3 } x + \frac { 22 } { 3 }$$ The region \(R\), shown shaded in Figure 3, is bounded by \(C _ { 1 } , l\) and the \(y\)-axis.
4. Find the exact area of \(R\).

9. \begin{figure}[h] \end{figure} In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 2 } { 3 } x ^ { 2 } - 9 \sqrt { x } + 13 \quad x \geqslant 0$$

1. Find, using calculus, the range of values of \(x\) for which \(y\) is increasing. The point \(P\) lies on \(C\) and has coordinates (9, 40).
   The line \(l\) is the tangent to \(C\) at the point \(P\).
   The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the line \(l\), the \(x\)-axis and the \(y\)-axis.
2. Find, using calculus, the exact area of \(R\).

8. Figure 2 Figure 2 shows a sketch of part of the curves \(C _ { 1 }\) and \(C _ { 2 }\) with equations $$\begin{array} { l l } C _ { 1 } : y = 10 x - x ^ { 2 } - 8 & x > 0 \\ C _ { 2 } : y = x ^ { 3 } & x > 0 \end{array}$$ The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the points \(A\) and \(B\).

1. Verify that the point \(A\) has coordinates (1, 1)
2. Use algebra to find the coordinates of the point \(B\) The finite region \(R\) is bounded by \(C _ { 1 }\) and \(C _ { 2 }\)
3. Use calculus to find the exact area of \(R\) \includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-23_936_759_118_582} \includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-26_2674_1948_107_118}

8. \begin{figure}[h] \end{figure} The line with equation \(y = 3 x + 20\) cuts the curve with equation \(y = x ^ { 2 } + 6 x + 10\) at the points \(A\) and \(B\), as shown in Figure 2.

1. Use algebra to find the coordinates of \(A\) and the coordinates of \(B\). The shaded region \(S\) is bounded by the line and the curve, as shown in Figure 2.
2. Use calculus to find the exact area of \(S\).

5. \begin{figure}[h] \end{figure} In Figure \(2 O A B\) is a sector of a circle radius 5 m . The chord \(A B\) is 6 m long.

1. Show that \(\cos A \hat { O } B = \frac { 7 } { 25 }\).
2. Hence find the angle \(A \hat { O } B\) in radians, giving your answer to 3 decimal places.
3. Calculate the area of the sector \(O A B\).
4. Hence calculate the shaded area.

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x ( x - 1 ) ( x - 5 )$$ Use calculus to find the total area of the finite region, shown shaded in Figure 1, that is between \(x = 0\) and \(x = 2\) and is bounded by \(C\), the \(x\)-axis and the line \(x = 2\).
(9)

7. \begin{figure}[h] \end{figure} The curve \(C\) has equation \(y = x ^ { 2 } - 5 x + 4\). It cuts the \(x\)-axis at the points \(L\) and \(M\) as shown in Figure 2.

1. Find the coordinates of the point \(L\) and the point \(M\).
2. Show that the point \(N ( 5,4 )\) lies on \(C\).
3. Find \(\int \left( x ^ { 2 } - 5 x + 4 \right) \mathrm { d } x\). The finite region \(R\) is bounded by \(L N , L M\) and the curve \(C\) as shown in Figure 2.
4. Use your answer to part (c) to find the exact value of the area of \(R\).
   \section*{LU}

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = ( x + 1 ) ( x - 5 )$$ The curve crosses the \(x\)-axis at the points \(A\) and \(B\).

1. Write down the \(x\)-coordinates of \(A\) and \(B\). The finite region \(R\), shown shaded in Figure 1, is bounded by \(C\) and the \(x\)-axis.
2. Use integration to find the area of \(R\).

6. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = \frac { 16 } { x ^ { 2 } } - \frac { x } { 2 } + 1 , \quad x > 0$$ The finite region \(R\), bounded by the lines \(x = 1\), the \(x\)-axis and the curve, is shown shaded in Figure 1. The curve crosses the \(x\)-axis at the point \(( 4,0 )\).

1. Complete the table with the values of \(y\) corresponding to \(x = 2\) and 2.5

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	16.5	7.361			1.278	0.556	0

2. Use the trapezium rule with all the values in the completed table to find an approximate value for the area of \(R\), giving your answer to 2 decimal places.
3. Use integration to find the exact value for the area of \(R\).

9. \(y\) \begin{figure}[h] \end{figure} The finite region \(R\), as shown in Figure 2, is bounded by the \(x\)-axis and the curve with equation $$y = 27 - 2 x - 9 \sqrt { } x - \frac { 16 } { x ^ { 2 } } , \quad x > 0$$ The curve crosses the \(x\)-axis at the points \(( 1,0 )\) and \(( 4,0 )\).

1. Complete the table below, by giving your values of \(y\) to 3 decimal places.

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	0	5.866		5.210		1.856	0

2. Use the trapezium rule with all the values in the completed table to find an approximate value for the area of \(R\), giving your answer to 2 decimal places.
3. Use integration to find the exact value for the area of \(R\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x ^ { 3 } - 6 x ^ { 2 } + 9 x + 5$$ The point \(P ( 4,9 )\) lies on \(C\).

1. Show that the normal to \(C\) at the point \(P\) has equation $$x + 9 y = 85$$ The region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the \(y\)-axis and the normal to \(C\) at \(P\).
2. Showing all your working, calculate the exact area of \(R\).

10. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve \(C\) with equation \(y = 2 x + \frac { 8 } { x ^ { 2 } } - 5 , x > 0\).
The points \(P\) and \(Q\) lie on \(C\) and have \(x\)-coordinates 1 and 4 respectively. The region \(R\), shaded in Figure 1, is bounded by \(C\) and the straight line joining \(P\) and \(Q\).

1. Find the exact area of \(R\).
2. Use calculus to show that \(y\) is increasing for \(x > 2\).

10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = x ^ { 3 } - 8 x ^ { 2 } + 20 x\). The curve has stationary points \(A\) and \(B\).

1. Use calculus to find the \(x\)-coordinates of \(A\) and \(B\).
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at \(A\), and hence verify that \(A\) is a maximum. The line through \(B\) parallel to the \(y\)-axis meets the \(x\)-axis at the point \(N\).
   The region \(R\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line from \(A\) to \(N\).
3. Find \(\int \left( x ^ { 3 } - 8 x ^ { 2 } + 20 x \right) \mathrm { d } x\).
4. Hence calculate the exact area of \(R\).

5. The curve \(C\) has equation $$y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , \quad 0 \leqslant x \leqslant 2$$

1. Complete the table below, giving the values of \(y\) to 3 decimal places at \(x = 1\) and \(x = 1.5\).

   \(x\)	0	0.5	1	1.5	2
   \(y\)	0	0.530			6

2. Use the trapezium rule, with all the \(y\) values from your table, to find an approximation for the value of \(\int _ { 0 } ^ { 2 } x \sqrt { } \left( x ^ { 3 } + 1 \right) \mathrm { d } x\), giving your answer to 3 significant figures. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{22ebc302-765c-4734-b312-b286ccb20be9-06_1110_644_1119_648} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows the curve \(C\) with equation \(y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , 0 \leqslant x \leqslant 2\), and the straight line segment \(l\), which joins the origin and the point \(( 2,6 )\). The finite region \(R\) is bounded by \(C\) and \(l\).
3. Use your answer to part (b) to find an approximation for the area of \(R\), giving your answer to 3 significant figures.
   (3) \section*{LU}

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = 10 + 8 x + x ^ { 2 } - x ^ { 3 }\).
The curve has a maximum turning point \(A\).

1. Using calculus, show that the \(x\)-coordinate of \(A\) is 2 . The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(y\)-axis and the line from \(O\) to \(A\), where \(O\) is the origin.
2. Using calculus, find the exact area of \(R\).

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = x ^ { 3 } - 10 x ^ { 2 } + k x$$ where \(k\) is a constant. The point \(P\) on \(C\) is the maximum turning point.
Given that the \(x\)-coordinate of \(P\) is 2 ,

1. show that \(k = 28\). The line through \(P\) parallel to the \(x\)-axis cuts the \(y\)-axis at the point \(N\). The region \(R\) is bounded by \(C\), the \(y\)-axis and \(P N\), as shown shaded in Figure 2.
2. Use calculus to find the exact area of \(R\).

9. \begin{figure}[h] \end{figure} The straight line with equation \(y = x + 4\) cuts the curve with equation \(y = - x ^ { 2 } + 2 x + 24\) at the points \(A\) and \(B\), as shown in Figure 3.

1. Use algebra to find the coordinates of the points \(A\) and \(B\). The finite region \(R\) is bounded by the straight line and the curve and is shown shaded in Figure 3.
2. Use calculus to find the exact area of \(R\).