Area under curve using integration

11
\includegraphics[max width=\textwidth, alt={}, center]{aaba3158-b5be-464e-bea3-1a4c460f9637-16_622_1091_260_525} The diagram shows part of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } + k ^ { 2 } x ^ { - \frac { 1 } { 2 } }\), where \(k\) is a positive constant.

1. Find the coordinates of the minimum point of the curve, giving your answer in terms of \(k\).
   The tangent at the point on the curve where \(x = 4 k ^ { 2 }\) intersects the \(y\)-axis at \(P\).
2. Find the \(y\)-coordinate of \(P\) in terms of \(k\).
   The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = \frac { 9 } { 4 } k ^ { 2 }\) and \(x = 4 k ^ { 2 }\).
3. Find the area of the shaded region in terms of \(k\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

6 The curve with equation \(y = 2 x - 8 x ^ { \frac { 1 } { 2 } }\) has a minimum point at \(A\) and intersects the positive \(x\)-axis at \(B\).

1. Find the coordinates of \(A\) and \(B\).
2. 
   \includegraphics[max width=\textwidth, alt={}, center]{84b3cd80-faf0-4522-8286-52bf7f86dc8a-11_522_1561_296_244} The diagram shows the curve with equation \(\mathrm { y } = 2 \mathrm { x } - 8 \mathrm { x } ^ { \frac { 1 } { 2 } }\) and the line \(A B\). It is given that the equation of \(A B\) is \(y = \frac { 2 x - 32 } { 3 }\). Find the area of the shaded region between the curve and the line.

10
\includegraphics[max width=\textwidth, alt={}, center]{c8925c7a-cb3b-43b8-9d09-8adc800c6887-4_798_805_258_669} The diagram shows part of the curve \(y = \frac { 8 } { \sqrt { } ( 3 x + 4 ) }\). The curve intersects the \(y\)-axis at \(A ( 0,4 )\). The normal to the curve at \(A\) intersects the line \(x = 4\) at the point \(B\).

1. Find the coordinates of \(B\).
2. Show, with all necessary working, that the areas of the regions marked \(P\) and \(Q\) are equal.

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

1. Find the coordinates of the stationary point on the curve and determine its nature.
2. Find the area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 1\).

3
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_776_483_310_769} The diagram shows the curve with equation \(y = 8 \mathrm { e } ^ { - x } - \mathrm { e } ^ { 2 x }\). The curve crosses the \(y\)-axis at the point \(A\) and the \(x\)-axis at the point \(B\). The shaded region is bounded by the curve and the two axes.

1. Find the gradient of the curve at \(A\).
   \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_2715_35_141_2011}
2. Show that the \(x\)-coordinate of \(B\) is \(\ln 2\) and hence find the area of the shaded region.

3
\includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-2_488_664_863_737} The diagram shows the curve \(y = \mathrm { e } ^ { 2 x }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 0 , y = 0\) and \(x = p\).

1. Find, in terms of \(p\), the area of \(R\).
2. Hence calculate the value of \(p\) for which the area of \(R\) is equal to 5 . Give your answer correct to 2 significant figures.

5
\includegraphics[max width=\textwidth, alt={}, center]{beb8df77-e091-4248-812b-20e885c42e37-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).

5
\includegraphics[max width=\textwidth, alt={}, center]{0a45a806-007f-4840-85e7-16d4c1a2c599-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).

7

1. Find the exact area of the region bounded by the curve \(y = 1 + \mathrm { e } ^ { 2 x - 1 }\), the \(x\)-axis and the lines \(x = \frac { 1 } { 2 }\) and \(x = 2\).
2. 
   \includegraphics[max width=\textwidth, alt={}, center]{e3ee4932-8219-4332-9cd2-e7f835522469-3_469_719_397_753} The diagram shows the curve \(y = \frac { \mathrm { e } ^ { 2 x } } { \sin 2 x }\) for \(0 < x < \frac { 1 } { 2 } \pi\), and its minimum point \(M\). Find the exact \(x\)-coordinate of \(M\).

4
\includegraphics[max width=\textwidth, alt={}, center]{cc051d68-7e21-4dc1-b34d-6fb7f12a52fd-2_524_625_1425_758} The diagram shows the curve \(y = \mathrm { e } ^ { x } + 4 \mathrm { e } ^ { - 2 x }\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) is \(\ln 2\).
2. The region shaded in the diagram is enclosed by the curve and the lines \(x = 0 , x = \ln 2\) and \(y = 0\). Use integration to show that the area of the shaded region is \(\frac { 5 } { 2 }\).

4
\includegraphics[max width=\textwidth, alt={}, center]{3b217eb4-3bd3-4800-a913-749754bf109f-2_524_625_1425_758} The diagram shows the curve \(y = \mathrm { e } ^ { x } + 4 \mathrm { e } ^ { - 2 x }\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) is \(\ln 2\).
2. The region shaded in the diagram is enclosed by the curve and the lines \(x = 0 , x = \ln 2\) and \(y = 0\). Use integration to show that the area of the shaded region is \(\frac { 5 } { 2 }\).

5
\includegraphics[max width=\textwidth, alt={}, center]{c5ec981d-7ff7-4698-82c4-eb0506b635a3-2_515_1031_1384_555} The diagram shows the curve \(y = \mathrm { e } ^ { - x } - \mathrm { e } ^ { - 2 x }\) and its maximum point \(M\). The \(x\)-coordinate of \(M\) is denoted by \(p\).

1. Find the exact value of \(p\).
2. Show that the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = p\) is equal to \(\frac { 1 } { 8 }\).

6

1. Find the value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } ( \sin 2 x + \cos x ) \mathrm { d } x\).
2. 
   \includegraphics[max width=\textwidth, alt={}, center]{9894d97f-3b7b-4dbe-b94a-2c8415442038-3_517_880_422_669} The diagram shows part of the curve \(y = \frac { 1 } { x + 1 }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 1 , y = 0\) and \(x = p\).
   1. Find, in terms of \(p\), the area of \(R\).
   2. Hence find, correct to 1 decimal place, the value of \(p\) for which the area of \(R\) is equal to 2 .

6
\includegraphics[max width=\textwidth, alt={}, center]{4029c46c-50a1-4d23-bc29-589417a6b7f5-3_501_497_269_826} The diagram shows the part of the curve \(y = \frac { \mathrm { e } ^ { 2 x } } { x }\) for \(x > 0\), and its minimum point \(M\).

1. Find the coordinates of \(M\).
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 1 } ^ { 2 } \frac { \mathrm { e } ^ { 2 x } } { x } \mathrm {~d} x$$ giving your answer correct to 1 decimal place.
3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).
4. Given that \(y = \tan 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
5. Hence, or otherwise, show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sec ^ { 2 } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \sqrt { } 3$$ and, by using an appropriate trigonometrical identity, find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \tan ^ { 2 } 2 x \mathrm {~d} x\).
6. Use the identity \(\cos 4 x \equiv 2 \cos ^ { 2 } 2 x - 1\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \frac { 1 } { 1 + \cos 4 x } \mathrm {~d} x$$

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 \(\int _ { 2 } ^ { 6 } \left( \frac { 36 } { x ^ { 2 } } + 2 x + k \right) \mathrm { d } x = 0\)

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 $$\int _ { - 3 } ^ { 6 } \left( \frac { 1 } { 2 } x ^ { 2 } + k \right) \mathrm { d } x = 55$$

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 \(x = 2\).

6．（a）Starting from \([ \mathrm { f } ( x ) - \lambda \mathrm { g } ( x ) ] ^ { 2 } \geqslant 0\) show that \(\lambda\) satisfies the quadratic inequality $$\left( \int _ { a } ^ { b } [ \operatorname { g } ( x ) ] ^ { 2 } \mathrm {~d} x \right) \lambda ^ { 2 } - 2 \left( \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { g } ( x ) \mathrm { d } x \right) \lambda + \int _ { a } ^ { b } [ \mathrm { f } ( x ) ] ^ { 2 } \mathrm {~d} x \geqslant 0$$ where \(a\) and \(b\) are constants and \(\lambda\) can take any real value．
（2）
（b）Hence prove that $$\left[ \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { g } ( x ) \mathrm { d } x \right] ^ { 2 } \leqslant \left[ \int _ { a } ^ { b } [ \mathrm { f } ( x ) ] ^ { 2 } \mathrm {~d} x \right] \times \left[ \int _ { a } ^ { b } [ \mathrm {~g} ( x ) ] ^ { 2 } \mathrm {~d} x \right]$$ （c）By letting \(\mathrm { f } ( x ) = 1\) and \(\mathrm { g } ( x ) = \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } }\) show that $$\int _ { - 1 } ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x \leqslant \frac { 9 } { 2 }$$ （d）Show that \(\int _ { - 1 } ^ { 2 } x ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 4 } } \mathrm {~d} x = \frac { 12 \sqrt { } 3 } { 5 }\)
（e）Hence show that $$\frac { 144 } { 55 } \leqslant \int _ { - 1 } ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$$

7. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve \(C\) with equation \(y = x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - 34 x\) and the line \(L\) with equation \(y = m x + c\) ． The line \(L\) touches \(C\) at the points \(P\) and \(Q\) with \(x\) coordinates \(p\) and \(q\) respectively．
（a）Explain why $$x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - ( 34 + m ) x - c = ( x - p ) ^ { 2 } ( x - q ) ^ { 2 }$$ The finite region \(R\) ，shown shaded in Figure 3，is bounded by \(C\) and \(L\) ．
（b）Use integration by parts to show that the area of \(R\) is \(\frac { ( q - p ) ^ { 5 } } { 30 }\)
（c）Show that $$( x - p ) ^ { 2 } ( x - q ) ^ { 2 } = x ^ { 4 } - 2 ( p + q ) x ^ { 3 } + S x ^ { 2 } - T x + U$$ where \(S , T\) and \(U\) are expressions to be found in terms of \(p\) and \(q\) ．
（d）Using part（a）and part（c）find the value of \(p\) ，the value of \(q\) and the equation of \(L\) ．