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

9. \includegraphics[max width=\textwidth, alt={}, center]{33f9663f-26bb-445e-af6e-ca5ca927f7dd-3_638_757_1064_493} The diagram shows the curve with equation \(y = 5 + x - x ^ { 2 }\) and the normal to the curve at the point \(P ( 1,5 )\).

1. Find an equation for the normal to the curve at \(P\) in the form \(y = m x + c\).
2. Find the coordinates of the point \(Q\), where the normal to the curve at \(P\) intersects the curve again.
3. Show that the area of the shaded region bounded by the curve and the straight line \(P Q\) is \(\frac { 4 } { 3 }\).

9. \includegraphics[max width=\textwidth, alt={}, center]{e4afa57d-5be3-42a6-ab35-39b0fdcc1681-3_559_732_824_388} The diagram shows the curve with equation \(y = \mathrm { f } ( x )\) which crosses the \(x\)-axis at the origin and at the points \(A\) and \(B\). Given that $$f ^ { \prime } ( x ) = 4 - 6 x - 3 x ^ { 2 }$$

1. find an expression for \(y\) in terms of \(x\),
2. show that \(A\) has coordinates ( \(- 4,0\) ) and find the coordinates of \(B\),
3. find the total area of the two regions bounded by the curve and the \(x\)-axis.

9. The diagram shows the curve \(C\) with equation \(y = 3 x - 4 \sqrt { x } + 2\) and the tangent to \(C\) at the point \(A\). Given that \(A\) has \(x\)-coordinate 4,

1. show that the tangent to \(C\) at \(A\) has the equation \(y = 2 x - 2\). The shaded region is bounded by \(C\), the tangent to \(C\) at \(A\) and the \(y\)-axis.
2. Find the area of the shaded region.

7. \includegraphics[max width=\textwidth, alt={}, center]{e5d62032-84ad-4e0b-9b72-ccfd8f4dbac8-3_499_721_248_552} The diagram shows part of the curve \(y = \mathrm { f } ( x )\) where \(\mathrm { f } ( x ) = \frac { 1 - 8 x ^ { 3 } } { x ^ { 2 } } , x \neq 0\).

1. Solve the equation \(\mathrm { f } ( x ) = 0\).
2. Find \(\int \mathrm { f } ( x ) \mathrm { d } x\).
3. Find the area of the shaded region bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = 2\).

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 \(1 \leqslant t \leqslant 5\) ?

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 \(y = 12\).

2 Fig. 9 shows a sketch of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and the line \(y = 6 x + 24\). \begin{figure}[h] \end{figure}

1. Differentiate \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and hence find the \(x\)-coordinates of the turning points of the curve. Give your answers to 2 decimal places.
2. You are given that the line and the curve intersect when \(x = 0\) and when \(x = - 4\). Find algebraically the \(x\)-coordinate of the other point of intersection.
3. Use calculus to find the area of the region bounded by the curve and the line \(y = 6 x + 24\) for \(- 4 \leqslant x \leqslant 0\), shown shaded on Fig. 9.

2 A cubic curve has equation \(y = x ^ { 3 } - 3 x ^ { 2 } + 1\).

1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
2. Show that the tangent to the curve at the point where \(x = - 1\) has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the \(x\) - and \(y\)-axes is 8 square units.

4 The equation of a curve is \(y = 9 x ^ { 2 } - x ^ { 4 }\).

1. Show that the curve meets the \(x\)-axis at the origin and at \(x = \pm a\), stating the value of \(a\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). Hence show that the origin is a minimum point on the curve. Find the \(x\)-coordinates of the maximum points.
3. Use calculus to find the area of the region bounded by the curve and the \(x\)-axis between \(x = 0\) and \(x = a\), using the value you found for \(a\) in part (i).

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 \(y = 12\).

3. The curve \(C\) has the equation \(y = 2 \mathrm { e } ^ { x } - 6 \ln x\) and passes through the point \(P\) with \(x\)-coordinate 1.

1. Find an equation for the tangent to \(C\) at \(P\). The tangent to \(C\) at \(P\) meets the coordinate axes at the points \(Q\) and \(R\).
2. Show that the area of triangle \(O Q R\), where \(O\) is the origin, is \(\frac { 9 } { 3 - \mathrm { e } }\).

1 Fig. 12 is a sketch of the curve \(y = 2 x ^ { 2 } - 11 x + 12\). \begin{figure}[h] \end{figure}

1. Show that the curve intersects the \(x\)-axis at \(( 4,0 )\) and find the coordinates of the other point of intersection of the curve and the \(x\)-axis.
2. Find the equation of the normal to the curve at the point \(( 4,0 )\). Show also that the area of the triangle bounded by this normal and the axes is 1.6 units \({ } ^ { 2 }\).
3. Find the area of the region bounded by the curve and the \(x\)-axis.

3 A cubic curve has equation \(y = x ^ { 3 } - 3 x ^ { 2 } + 1\).

1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
2. Show that the tangent to the curve at the point where \(x = - 1\) has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the \(x\) - and \(y\)-axes is 8 square units.

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.

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.

1

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-1_650_759_252_762} \captionsetup{labelformat=empty} \caption{Fig. 12}
   \end{figure} Fig. 12 shows part of the curve \(y = x ^ { 4 }\) and the line \(y = 8 x\), which intersect at the origin and the point P .
   (A) Find the coordinates of P , and show that the area of triangle OPQ is 16 square units.
   (B) Find the area of the region bounded by the line and the curve.
2. You are given that \(\mathrm { f } ( x ) = x ^ { 4 }\).
   (A) Complete this identity for \(\mathrm { f } ( x + h )\). $$\mathrm { f } ( x + h ) = ( x + h ) ^ { 4 } = x ^ { 4 } + 4 x ^ { 3 } h + \ldots$$ (B) Simplify \(\frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
   (C) Find \(\lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
   (D) State what this limit represents.

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.

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

1. Find the exact area of \(R\). The region \(R\) is rotated completely about the \(x\)-axis.
2. Find the volume of the solid formed, giving your answer in terms of \(\pi\).

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.

5 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-2_486_746_1978_696} The diagram shows the curves \(y = ( 1 - 2 x ) ^ { 5 }\) and \(y = \mathrm { e } ^ { 2 x - 1 } - 1\). The curves meet at the point \(\left( \frac { 1 } { 2 } , 0 \right)\). Find the exact area of the region (shaded in the diagram) bounded by the \(y\)-axis and by part of each curve.

8 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-4_787_742_276_719} The diagram shows part of the curve \(y = \ln \left( 5 - x ^ { 2 } \right)\) which meets the \(x\)-axis at the point \(P\) with coordinates \(( 2,0 )\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(P Q\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\).
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures.
3. Deduce an approximation to the area of the region \(B\).

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 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-2_419_707_1576_660} The diagram shows the curve $$y = \frac { 1 } { \sqrt { } ( 4 x + 1 ) }$$ The region \(R\) (shaded in the diagram) is enclosed by the curve, the axes and the line \(x = 2\).

1. Show that the exact area of \(R\) is 1 .
2. The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed.