Stationary points then area/volume

9 \includegraphics[max width=\textwidth, alt={}, center]{3eefd6c1-924c-4b7e-8d17-a2942fb48234-3_399_696_255_721} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { 2 - x }\) and its maximum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) is 2 .
2. Find the exact value of \(\int _ { 0 } ^ { 2 } x ^ { 2 } \mathrm { e } ^ { 2 - x } \mathrm {~d} x\).

6 \includegraphics[max width=\textwidth, alt={}, center]{79efa364-da5a-4888-85a9-dc4de1e0908e-3_543_825_287_660} The diagram shows the curve \(y = ( 3 - x ) \mathrm { e } ^ { - 2 x }\) and its minimum point \(M\). The curve intersects the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

1. Calculate the \(x\)-coordinate of \(M\).
2. Find the area of the region bounded by \(O A , O B\) and the curve, giving your answer in terms of e.

7 \includegraphics[max width=\textwidth, alt={}, center]{f4614578-f5f6-4283-8185-8b5598ad91d5-3_416_679_258_731} The diagram shows part of the curve \(y = \left( 2 x - x ^ { 2 } \right) \mathrm { e } ^ { \frac { 1 } { 2 } x }\) and its maximum point \(M\).

1. Find the exact \(x\)-coordinate of \(M\).
2. Find the exact value of the area of the shaded region bounded by the curve and the positive \(x\)-axis.

7 \includegraphics[max width=\textwidth, alt={}, center]{84df6b9a-6118-44a2-9c18-512039ded4fd-3_416_677_258_733} The diagram shows part of the curve \(y = \left( 2 x - x ^ { 2 } \right) \mathrm { e } ^ { \frac { 1 } { 2 } x }\) and its maximum point \(M\).

1. Find the exact \(x\)-coordinate of \(M\).
2. Find the exact value of the area of the shaded region bounded by the curve and the positive \(x\)-axis.

9 \includegraphics[max width=\textwidth, alt={}, center]{be0fb208-2ef1-4fae-84ff-ad2e8bf2dcc5-16_446_956_260_593} The diagram shows the curve \(y = \left( 1 + x ^ { 2 } \right) \mathrm { e } ^ { - \frac { 1 } { 2 } x }\) for \(x \geqslant 0\). The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Find the exact values of the \(x\)-coordinates of the stationary points of the curve.
2. Show that the exact value of the area of \(R\) is \(18 - \frac { 42 } { \mathrm { e } }\).

9 \includegraphics[max width=\textwidth, alt={}, center]{21878d10-7f16-4dbb-86ef-65a7ba5eeafb-16_446_956_260_593} The diagram shows the curve \(y = \left( 1 + x ^ { 2 } \right) \mathrm { e } ^ { - \frac { 1 } { 2 } x }\) for \(x \geqslant 0\). The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Find the exact values of the \(x\)-coordinates of the stationary points of the curve.
2. Show that the exact value of the area of \(R\) is \(18 - \frac { 42 } { \mathrm { e } }\).

7 \includegraphics[max width=\textwidth, alt={}, center]{25dffd43-9456-449b-be77-8402109ee603-3_608_672_283_733} The diagram shows the curve \(y = 2 \mathrm { e } ^ { x } + 3 \mathrm { e } ^ { - 2 x }\). The curve cuts the \(y\)-axis at \(A\).

1. Write down the coordinates of \(A\).
2. Find the equation of the tangent to the curve at \(A\), and state the coordinates of the point where this tangent meets the \(x\)-axis.
3. Calculate the area of the region bounded by the curve and by the lines \(x = 0 , y = 0\) and \(x = 1\), giving your answer correct to 2 significant figures.

10 \includegraphics[max width=\textwidth, alt={}, center]{19aff1b7-51b7-4d44-86e6-45dad32aa121-16_426_908_262_616} The diagram shows the curve \(y = ( 2 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x }\), and its minimum point \(M\).

1. Find the exact coordinates of \(M\).
2. Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e.

9 \includegraphics[max width=\textwidth, alt={}, center]{39cf66af-095b-404b-a38c-0aa7684c4a27-14_428_787_274_671} The diagram shows the curve \(y = \sin x \cos 2 x\), for \(0 \leqslant x \leqslant \pi\), and a maximum point \(M\), where \(x = a\). The shaded region between the curve and the \(x\)-axis is denoted by \(R\).

1. Find the value of \(a\) correct to 2 decimal places.
2. Find the exact area of the region \(R\), giving your answer in simplified form.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\) The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\).

1. Find, in terms of \(\ln 2\), the \(x\) coordinate of the point \(A\).
2. Find $$\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the curve with equation $$y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$$
3. Find, by integration, the exact value for the area of \(R\). Give your answer in terms of \(\ln 2\)

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\) The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\) .

1. Find,in terms of \(\ln 2\) ,the \(x\) coordinate of the point \(A\) .
2. Find \(\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x\) The finite region \(R\) ,shown shaded in Figure 2,is bounded by the \(x\)-axis and the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\)
3. Find,by integration,the exact value for the area of \(R\) . Give your answer in terms of \(\ln 2\) \includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-18_2655_1943_114_118}

8

1. Given that \(\mathrm { e } ^ { - 2 x } = 3\), find the exact value of \(x\).
2. Use integration by parts to find \(\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x\).
3. A curve has equation \(y = \mathrm { e } ^ { - 2 x } + 6 x\).
   1. Find the exact values of the coordinates of the stationary point of the curve.
   2. Determine the nature of the stationary point.
   3. The region \(R\) is bounded by the curve \(y = \mathrm { e } ^ { - 2 x } + 6 x\), the \(x\)-axis and the lines \(x = 0\) and \(x = 1\). Find the volume of the solid formed when \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis, giving your answer to three significant figures.

7

1. A curve has equation \(y = x ^ { 2 } \mathrm { e } ^ { - \frac { x } { 4 } }\).
   Show that the curve has exactly two stationary points and find the exact values of their coordinates.
   (7 marks)
2. 1. Use integration by parts twice to find the exact value of \(\int _ { 0 } ^ { 4 } x ^ { 2 } \mathrm { e } ^ { - \frac { x } { 4 } } \mathrm {~d} x\).
   2. The region bounded by the curve \(y = 3 x \mathrm { e } ^ { - \frac { x } { 8 } }\), the \(x\)-axis from 0 to 4 and the line \(x = 4\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to form a solid. Use your answer to part (b)(i) to find the exact value of the volume of the solid generated.

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]{9aac4ee4-2435-4315-a87d-fe9fa8e15665-007_593_1034_696_543}
   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 $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$

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 $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$