Questions — OCR MEI C2 (480 questions)

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OCR MEI C2 Q11
4 marks Moderate -0.8
11 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - x ^ { 2 }\). The curve passes through the point \(( 6,1 )\). Find the equation of the curve.
OCR MEI C2 Q12
5 marks Easy -1.2
12 Given tha \(y = 6 x ^ { 3 } + \sqrt { x } + 3\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
OCR MEI C2 Q2
3 marks Easy -1.2
2 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = x ^ { 6 } + \sqrt { x }\).
  1. Find the equation of the tangent to the curve \(y = x ^ { 4 }\) at the point where \(x = 2\). Give your answer in the form \(y = m x + c\).
  2. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 4 }\) where \(x = 2\) and \(x = 2.1\).
  3. (A) Expand \(( 2 + h ) ^ { 4 }\).
    (B) Simplify \(\frac { ( 2 + h ) ^ { 4 } - 2 ^ { 4 } } { h }\).
    (C) Show how your result in part (iii) (B) can be used to find the gradient of \(y = x ^ { 4 }\) at the point where \(x = 2\).
  4. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 2 } - 7\) for which \(x = 3\) and \(x = 3.1\).
  5. Given that \(\mathrm { f } ( x ) = x ^ { 2 } - 7\), find and simplify \(\frac { \mathrm { f } ( 3 + h ) - \mathrm { f } ( 3 ) } { h }\).
  6. Use your result in part (ii) to find the gradient of \(y = x ^ { 2 } - 7\) at the point where \(x = 3\), showing your reasoning.
  7. Find the equation of the tangent to the curve \(y = x ^ { 2 } - 7\) at the point where \(x = 3\).
  8. This tangent crosses the \(x\)-axis at the point P . The curve crosses the positive \(x\)-axis at the point Q . Find the distance PQ , giving your answer correct to 3 decimal places.
  9. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8ff8b67d-1489-4cb1-bcd2-b32db674e29f-3_651_770_242_737} \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.
  10. 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.
OCR MEI C2 Q2
11 marks Moderate -0.3
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]
\includegraphics[alt={},max width=\textwidth]{4a9ca68f-f980-4a8f-b387-80dbdca33dfe-2_782_1168_319_451} \captionsetup{labelformat=empty} \caption{Fig. 9}
\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.
OCR MEI C2 Q3
11 marks Moderate -0.3
3
  1. The standard formulae for the volume \(V\) and total surface area \(A\) of a solid cylinder of radius \(r\) and height \(h\) are $$V = \pi r ^ { 2 } h \quad \text { and } \quad A = 2 \pi r ^ { 2 } + 2 \pi r h .$$ Use these to show that, for a cylinder with \(A = 200\), $$V = 100 r - \pi r ^ { 3 }$$
  2. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) and \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} r ^ { 2 } }\).
  3. Use calculus to find the value of \(r\) that gives a maximum value for \(V\) and hence find this maximum value, giving your answers correct to 3 significant figures.
OCR MEI C2 Q4
5 marks Moderate -0.8
4
  1. Differentiate \(x ^ { 3 } - 6 x ^ { 2 } - 15 x + 50\).
  2. Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 6 x ^ { 2 } - 15 x + 50\).
OCR MEI C2 Q5
5 marks Moderate -0.5
5 Use calculus to find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 6 x ^ { 2 } - 15 x\). Hence find the set of values of \(x\) for which \(x ^ { 3 } - 6 x ^ { 2 } - 15 x\) is an increasing function.
OCR MEI C2 Q1
12 marks Moderate -0.3
1 The equation of a cubic curve is \(y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the tangent to the curve when \(x = 3\) passes through the point \(( - 1 , - 41 )\).
  2. Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.
  3. Sketch the curve, given that the only real root of \(2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0\) is \(x = 0.2\) correct to 1 decimal place.
OCR MEI C2 Q2
13 marks Standard +0.3
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.
OCR MEI C2 Q3
5 marks Moderate -0.8
3 A curve has equation \(y = x + \frac { 1 } { x }\).
Use calculus to show that the curve has a turning point at \(x = 1\).
Show also that this point is a minimum.
OCR MEI C2 Q4
12 marks Moderate -0.8
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).
OCR MEI C2 Q5
5 marks Moderate -0.3
5 Differentiate \(4 x ^ { 2 } + \frac { 1 } { x }\) and hence find the \(x\)-coordinate of the stationary point of the curve \(y = 4 x ^ { 2 } + \frac { 1 } { x }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bba82ee6-90b2-4f03-9bb9-0371ff711a09-3_639_1027_302_542} \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.
OCR MEI C2 Q1
12 marks Moderate -0.3
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 \(y = 12\).
OCR MEI C2 Q2
13 marks Moderate -0.3
2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3\). The curve passes through the point ( 2,9 ).
  1. Find the equation of the tangent to the curve at the point \(( 2,9 )\).
  2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve.
  3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac { 1 } { 2 }\). Write down the coordinates of the minimum point of the transformed curve. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-2_1020_940_244_679} \captionsetup{labelformat=empty} \caption{Fig. 11}
    \end{figure} Fig. 11 shows a sketch of the cubic curve \(y = \mathrm { f } ( x )\). The values of \(x\) where it crosses the \(x\)-axis are - 5 , - 2 and 2 , and it crosses the \(y\)-axis at \(( 0 , - 20 )\).
  4. Express \(\mathrm { f } ( x )\) in factorised form.
  5. Show that the equation of the curve may be written as \(y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20\).
  6. Use calculus to show that, correct to 1 decimal place, the \(x\)-coordinate of the minimum point on the curve is 0.4 . Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.
  7. State, correct to 1 decimal place, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( 2 x )\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-3_768_1023_223_598} \captionsetup{labelformat=empty} \caption{Fig. 11}
    \end{figure} Fig. 11 shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).
  8. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
  9. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.
  10. Differentiate \(x ^ { 3 } - 3 x ^ { 2 } - 9 x\). Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 9 x\), showing which is the maximum and which the minimum.
  11. Find, in exact form, the coordinates of the points at which the curve crosses the \(x\)-axis.
  12. Sketch the curve. A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).
  13. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
  14. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).
OCR MEI C2 Q1
12 marks Standard +0.3
1 Fig. 12 is a sketch of the curve \(y = 2 x ^ { 2 } - 11 x + 12\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{44b860fb-040f-4d3f-94d8-42eae41c0e2d-1_468_940_285_830} \captionsetup{labelformat=empty} \caption{Fig. 12}
\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.
OCR MEI C2 Q2
11 marks Moderate -0.8
2 A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).
  1. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
  2. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).
OCR MEI C2 Q3
13 marks Standard +0.3
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.
OCR MEI C2 Q4
13 marks Moderate -0.3
4 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12e190fc-437f-499d-9c27-da49a7546755-2_604_912_1100_638} \captionsetup{labelformat=empty} \caption{Fig. 10}
\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 \(y = x ^ { 3 } - 6 x + 2\).
  4. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  5. Find, in exact form, the range of values of \(x\) for which \(x ^ { 3 } - 6 x + 2\) is a decreasing function.
  6. 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.
OCR MEI C2 Q1
10 marks Moderate -0.8
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]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-1_735_1246_335_441} \captionsetup{labelformat=empty} \caption{Fig. 12}
\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.
OCR MEI C2 Q2
4 marks Easy -1.2
2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-2_639_1037_294_517} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\) - and \(y\)-axes.
OCR MEI C2 Q3
12 marks Moderate -0.3
3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.
  1. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
    \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
  2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
    \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
    (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
    (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
OCR MEI C2 Q4
13 marks Moderate -0.3
4
  1. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_765_757_203_764} \captionsetup{labelformat=empty} \caption{Fig. 11.1}
    \end{figure} A boat travels from P to Q and then to R . As shown in Fig. 11.1, Q is 10.6 km from P on a bearing of \(045 ^ { \circ }\). R is 9.2 km from P on a bearing of \(113 ^ { \circ }\), so that angle QPR is \(68 ^ { \circ }\). Calculate the distance and bearing of R from Q .
  2. Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_527_1474_1452_404} \captionsetup{labelformat=empty} \caption{Fig. 11.2}
    \end{figure} BC is an arc of a circle with centre A and radius 80 cm . Angle \(\mathrm { CAB } = \frac { 2 \pi } { 3 }\) radians.
    EC is an arc of a circle with centre D and radius \(r \mathrm {~cm}\). Angle CDE is a right angle.
    1. Calculate the area of sector ABC .
    2. Show that \(r = 40 \sqrt { 3 }\) and calculate the area of triangle CDA.
    3. Hence calculate the area of cross-section of the rudder. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-5_695_1012_271_600} \captionsetup{labelformat=empty} \caption{Fig. 12}
      \end{figure} A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.
      1. Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough. Hence estimate the volume of water in the trough.
      2. A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation \(y = 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15\), for \(0 \leqslant x \leqslant 0.5\). Calculate \(\int _ { 0 } ^ { 0.5 } \left( 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15 \right) \mathrm { d } x\) and state what this represents.
        Hence find the volume of water in the trough as given by this model.
OCR MEI C2 Q1
12 marks Moderate -0.3
1 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10 -second intervals. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-1_732_753_302_700} \captionsetup{labelformat=empty} \caption{Fig. 10}
\end{figure} The measured speeds are shown below.
Time \(( t\) seconds \()\)0102030405060
Speed \(\left( v \mathrm {~m} \mathrm {~s} ^ { 1 } \right)\)28191411121622
  1. Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line \(t = 60\) and the axes. [This area represents the distance travelled by the car.]
  2. Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by \(v = 28 - t + 0.015 t ^ { 2 }\).
  3. Show that the difference between the value given by the model when \(t = 10\) and the measured value is less than \(3 \%\) of the measured value.
  4. According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 \quad t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.
OCR MEI C2 Q2
3 marks Easy -1.2
2 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results.
Distance across river \(( \mathrm { m } )\)01.534.567.5
Depth of river \(( \mathrm { m } )\)0.62.33.12.81.80.7
Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.
OCR MEI C2 Q3
12 marks Moderate -0.8
3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-2_579_1385_1035_424} \captionsetup{labelformat=empty} \caption{Fig. 11}
\end{figure}
  1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
  2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. The curve of the roof may be modelled by \(y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.
  3. Use integration to find the area of the cross-section according to this model.
  4. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\).