1.08b Integrate x^n: where n != -1 and sums

9 \includegraphics[max width=\textwidth, alt={}, center]{9e95415c-00f5-4b52-a443-0b946602b3b4-4_387_624_287_717} The diagram shows part of the curve \(y = - 3 + 2 \sqrt { x + 4 }\). The point \(P ( 5,3 )\) lies on the curve. Region \(A\) is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 5\). Region \(B\) is bounded by the curve, the \(y\)-axis and the line \(y = 3\).

1. Use the trapezium rule, with 2 strips each of width 2.5 , to find an approximate value for the area of region \(A\), giving your answer correct to 3 significant figures.
2. Use your answer to part (i) to deduce an approximate value for the area of region \(B\).
3. By first writing the equation of the curve in the form \(x = \mathrm { f } ( y )\), use integration to show that the exact area of region \(B\) is \(\frac { 14 } { 3 }\). \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {N } }\)}

5 A curve has an equation which satisfies \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 3 x ^ { - \frac { 1 } { 2 } }\) for all positive values of \(x\). The point \(P ( 4,1 )\) lies on the curve, and the gradient of the curve at \(P\) is 5 . Find the equation of the curve.

6 The cubic polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = x ^ { 3 } - 19 x + 30\).

1. Given that \(x = 2\) is a root of the equation \(\mathrm { f } ( x ) = 0\), express \(\mathrm { f } ( x )\) as the product of 3 linear factors.
2. Use integration to find the exact value of \(\int _ { - 5 } ^ { 3 } \mathrm { f } ( x ) \mathrm { d } x\).
3. Explain with the aid of a sketch why the answer to part (ii) does not give the area enclosed by the curve \(y = \mathrm { f } ( x )\) and the \(x\)-axis for \(- 5 \leqslant x \leqslant 3\).

5

1. Find \(\int \left( x ^ { 2 } + 2 \right) ( 2 x - 3 ) \mathrm { d } x\).
2. 1. Find, in terms of \(a\), the value of \(\int _ { 1 } ^ { a } \left( 6 x ^ { - 2 } - 4 x ^ { - 3 } \right) \mathrm { d } x\), where \(a\) is a constant greater than 1 .
   2. Deduce the value of \(\int _ { 1 } ^ { \infty } \left( 6 x ^ { - 2 } - 4 x ^ { - 3 } \right) \mathrm { d } x\).

1 Find \(\int \left( 20 x ^ { 4 } + 6 x ^ { - \frac { 3 } { 2 } } \right) \mathrm { d } x\).
[0pt] [4]

2 Find \(\int \left( 3 x ^ { 5 } + 2 x ^ { - \frac { 1 } { 2 } } \right) \mathrm { d } x\).

2 Find \(\int \left( x ^ { 5 } + 10 x ^ { \frac { 3 } { 2 } } \right) \mathrm { d } x\).

6 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \sqrt { x } - 2\). Given also that the curve passes through the point \(( 9,4 )\), find the equation of the curve.

7 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { \frac { 1 } { 2 } } - 5\). Given also that the curve passes through the point (4, 20), find the equation of the curve.

9 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]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_437_640_470_715} \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]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_574_808_1402_632} \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.

1

1. Differentiate \(12 \sqrt [ 3 ] { x }\).
2. Integrate \(\frac { 6 } { x ^ { 3 } }\).

2 Find

1. \(\int \left( 2 - \frac { 1 } { x } \right) ^ { 2 } \mathrm {~d} x\),
2. \(\int ( 4 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\).

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), which has a \(y\)-intercept at \(\mathrm { P } ( 0,3 )\), a minimum point at \(\mathrm { Q } ( 1,2 )\), and an asymptote \(x = - 1\). \begin{figure}[h] \end{figure}

1. Find the coordinates of the images of the points P and Q when the curve \(y = \mathrm { f } ( x )\) is transformed to
   (A) \(y = 2 \mathrm { f } ( x )\),
   (B) \(y = \mathrm { f } ( x + 1 ) + 2\). You are now given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1\).
2. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence find the coordinates of the other turning point on the curve \(y = \mathrm { f } ( x )\).
3. Show that \(\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }\).
4. Find \(\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x\) in terms of \(a\) and \(b\). Hence, by choosing suitable values for \(a\) and \(b\), find the exact area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\).

9

1. Find \(\int ( x + \cos 2 x ) ^ { 2 } \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{80f94db1-39be-46f5-896e-277c93cbe4b8-3_538_935_383_646} The diagram shows the part of the curve \(y = x + \cos 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The shaded region bounded by the curve, the axes and the line \(x = \frac { 1 } { 2 } \pi\) is rotated completely about the \(x\)-axis to form a solid of revolution of volume \(V\). Find \(V\), giving your answer in an exact form.

3

1. Find the quotient when \(3 x ^ { 3 } - x ^ { 2 } + 10 x - 3\) is divided by \(x ^ { 2 } + 3\), and show that the remainder is \(x\).
2. Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 3 x ^ { 3 } - x ^ { 2 } + 10 x - 3 } { x ^ { 2 } + 3 } \mathrm {~d} x$$

6 A particle moves along a straight line through an origin O . Initially the particle is at O .
At time \(t \mathrm {~s}\), its displacement from O is \(x \mathrm {~m}\) and its velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), is given by $$v = 24 - 18 t + 3 t ^ { 2 } .$$

1. Find the times, \(T _ { 1 } \mathrm {~s}\) and \(T _ { 2 } \mathrm {~s}\) (where \(T _ { 2 } > T _ { 1 }\) ), at which the particle is stationary.
2. Find an expression for \(x\) at time \(t \mathrm {~s}\). Show that when \(t = T _ { 1 } , x = 20\) and find the value of \(x\) when \(t = T _ { 2 }\). Section B (36 marks) \(7 \quad\) Abi and Bob are standing on the ground and are trying to raise a small object of weight 250 N to the top of a building. They are using long light ropes. Fig. 7.1 shows the initial situation. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-4_773_1071_429_497} \captionsetup{labelformat=empty} \caption{Fig. 7.1}
   \end{figure} Abi pulls vertically downwards on the rope A with a force \(F\) N. This rope passes over a small smooth pulley and is then connected to the object. Bob pulls on another rope, B, in order to keep the object away from the side of the building. In this situation, the object is stationary and in equilibrium. The tension in rope B, which is horizontal, is 25 N . The pulley is 30 m above the object. The part of rope A between the pulley and the object makes an angle \(\theta\) with the vertical.

5 Use de Moivre's theorem to express \(\cos ^ { 4 } \theta\) in the form $$a \cos 4 \theta + b \cos 2 \theta + c$$ where \(a , b , c\) are constants to be found. Hence evaluate $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 4 } \theta d \theta$$ leaving your answer in terms of \(\pi\).

10. $$g ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 41 x - 70$$

1. Use the factor theorem to show that \(\mathrm { g } ( x )\) is divisible by \(( x - 5 )\).
2. Hence, showing all your working, write \(\mathrm { g } ( x )\) as a product of three linear factors. The finite region \(R\) is bounded by the curve with equation \(y = \mathrm { g } ( x )\) and the \(x\)-axis, and lies below the \(x\)-axis.
3. Find, using algebraic integration, the exact value of the area of \(R\).

1. Find

$$\int \left( 8 x ^ { 3 } - \frac { 3 } { 2 \sqrt { x } } + 5 \right) \mathrm { d } x$$ giving your answer in simplest form.

1. Find

$$\int \frac { 2 \sqrt { x } - 3 } { x ^ { 2 } } \mathrm {~d} x$$ giving your answer in simplest form.

1. Find

$$\int \frac { 3 x ^ { 4 } - 4 } { 2 x ^ { 3 } } d x$$ writing your answer in simplest form.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \sqrt { x } , x \geqslant 0\) The region \(R\), shown shaded in Figure 2, is bounded by the curve, the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = a\), where \(a\) is a constant. Given that the area of \(R\) is 10

1. find, in simplest form, the value of
   1. \(\int _ { 1 } ^ { a } \sqrt { 8 x } \mathrm {~d} x\)
   2. \(\int _ { 0 } ^ { a } \sqrt { x } \mathrm {~d} x\)
2. show that \(a = 2 ^ { k }\), where \(k\) is a rational constant to be found.

1. Find

$$\int \frac { x ^ { \frac { 1 } { 2 } } ( 2 x - 5 ) } { 3 } \mathrm {~d} x$$ writing each term in simplest form.