Integration with differentiation context

10 \includegraphics[max width=\textwidth, alt={}, center]{6bcc553c-4938-46ef-bba4-97391b4d58d4-14_378_666_264_737} The diagram shows part of the curve \(y = \frac { 2 } { ( 3 - 2 x ) ^ { 2 } } - x\) and its minimum point \(M\), which lies on the \(x\)-axis.

1. Find expressions for \(\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\int y \mathrm {~d} x\).
2. Find, by calculation, the \(x\)-coordinate of \(M\).
3. Find the area of the shaded region bounded by the curve and the coordinate axes.

11 \includegraphics[max width=\textwidth, alt={}, center]{88c7a3f3-e129-4e9c-acf8-8c96d2668d43-14_693_782_267_669} The diagram shows part of the curve with equation \(y = x + \frac { 2 } { ( 2 x - 1 ) ^ { 2 } }\). The lines \(x = 1\) and \(x = 2\) intersect the curve at \(P\) and \(Q\) respectively and \(R\) is the stationary point on the curve.

1. Verify that the \(x\)-coordinate of \(R\) is \(\frac { 3 } { 2 }\) and find the \(y\)-coordinate of \(R\).
2. Find the exact value of the area of the shaded region.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 \includegraphics[max width=\textwidth, alt={}, center]{8c533469-393c-4e4c-a6ec-eab1303741e7-3_480_901_973_621} The diagram shows the curve \(y = x ^ { 2 } e ^ { - \frac { 1 } { 2 } x }\).

1. Find the \(x\)-coordinate of \(M\), the maximum point of the curve.
2. Find the area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = 1\), giving your answer in terms of e.

6 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-08_318_750_260_699} The diagram shows the curve \(y = \frac { x } { 1 + 3 x ^ { 4 } }\), for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
2. Using the substitution \(u = \sqrt { 3 } x ^ { 2 }\), find by integration the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 1\).

9 The equation of a curve is \(y = x ^ { - \frac { 2 } { 3 } } \ln x\) for \(x > 0\). The curve has one stationary point.

1. Find the exact coordinates of the stationary point.
2. Show that \(\int _ { 1 } ^ { 8 } y \mathrm {~d} x = 18 \ln 2 - 9\).

8 \includegraphics[max width=\textwidth, alt={}, center]{1990cbac-d96f-4484-be4b-67dab35b3147-12_458_725_262_708} The diagram shows the curve \(y = \frac { \ln x } { x ^ { 4 } }\) and its maximum point \(M\).

1. Find the exact coordinates of \(M\).
2. By using integration by parts, show that for all \(a > 1 , \int _ { 1 } ^ { a } \frac { \ln x } { x ^ { 4 } } \mathrm {~d} x < \frac { 1 } { 9 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{5eb2657c-ed74-4ed2-b8c4-08e9e0f657b5-08_351_1031_264_516} The diagram shows the curve \(\mathrm { y } = \mathrm { xe } ^ { - \mathrm { ax } }\), where \(a\) is a positive constant, and its maximum point \(M\).

1. Find the exact coordinates of \(M\).
2. Find the exact value of \(\int _ { 0 } ^ { \frac { 2 } { a } } x e ^ { - a x } d x\).

8 \includegraphics[max width=\textwidth, alt={}, center]{8c26235b-c78c-40d8-9e8e-213dc1311186-12_437_686_274_719} The diagram shows the curve \(y = x ^ { 3 } \ln x\), for \(x > 0\), and its minimum point \(M\).

1. Find the exact coordinates of \(M\).
2. Find the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = \frac { 1 } { 2 }\). [5]

10 \includegraphics[max width=\textwidth, alt={}, center]{5f80ae11-34c3-4d2f-89f8-71b4ac021c7d-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]{98001cfe-46a1-4c8f-9230-c140ebff6176-14_535_1082_274_520} The diagram shows part of the curve \(y = ( 3 - x ) \mathrm { e } ^ { - \frac { 1 } { 3 } x }\) for \(x \geqslant 0\), 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, giving your answer in terms of e.

10 \includegraphics[max width=\textwidth, alt={}, center]{a49b720b-f8d2-42ff-b147-5d676993aa4c-16_611_689_274_721} The diagram shows the curve \(y = x \cos 2 x\), for \(x \geqslant 0\).

1. Find the equation of the tangent to the curve at the point where \(x = \frac { 1 } { 2 } \pi\).
2. Find the exact area of the shaded region shown in the diagram, bounded by the curve and the \(x\)-axis.

7 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

9. \begin{figure}[h] \end{figure} Figure 5 shows a sketch of part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 6 x ^ { 2 } + 4 x - 2 } { 2 x + 1 } \quad x > - \frac { 1 } { 2 }$$

1. Find \(\mathrm { f } ^ { \prime } ( x )\), giving the answer in simplest form. The line \(l\) is the normal to \(C\) at the point \(P ( 2,6 )\)
2. Show that an equation for \(l\) is $$16 y + 5 x = 106$$
3. Write \(\mathrm { f } ( x )\) in the form \(A x + B + \frac { D } { 2 x + 1 }\) where \(A , B\) and \(D\) are constants. The region \(R\), shown shaded in Figure 5, is bounded by \(C , l\) and the \(x\)-axis.
4. Use algebraic integration to find the exact area of \(R\), giving your answer in the form \(P + Q \ln 3\), where \(P\) and \(Q\) are rational constants.
   (Solutions based entirely on calculator technology are not acceptable.)

8

1. Using the quotient rule, show that \(\frac { \mathrm { d } } { \mathrm { d } \theta } ( \tan 3 \theta ) = 3 + 3 \tan ^ { 2 } 3 \theta\) for \(- \frac { 1 } { 6 } \pi < \theta < \frac { 1 } { 6 } \pi\).
2. Hence find the value of \(\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 9 } \pi } \tan ^ { 2 } 3 \theta \mathrm {~d} \theta\), giving your answer in the simplest exact form.

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{\cos^2 x}\), \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), together with its asymptotes \(x = \frac{1}{2}\pi\) and \(x = -\frac{1}{2}\pi\). \includegraphics{figure_9}

1. Use the quotient rule to show that the derivative of \(\frac{\sin x}{\cos x}\) is \(\frac{1}{\cos^2 x}\). [3]
2. Find the area bounded by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\pi\). [3]

The function \(g(x)\) is defined by \(g(x) = \frac{1}{2}f(x + \frac{1}{4}\pi)\).

1. Verify that the curves \(y = f(x)\) and \(y = g(x)\) cross at \((0, 1)\). [3]
2. State a sequence of two transformations such that the curve \(y = f(x)\) is mapped to the curve \(y = g(x)\). On the copy of Fig. 9, sketch the curve \(y = g(x)\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve. [8]
3. Use your result from part (ii) to write down the area bounded by the curve \(y = g(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = -\frac{1}{4}\pi\). [1]