1.08d Evaluate definite integrals: between limits

5 In this question, you are required to investigate the curve with equation $$y = x ^ { m } ( 1 - x ) ^ { n } , \quad 0 \leqslant x \leqslant 1 ,$$ for various positive values of \(m\) and \(n\).

1. On separate diagrams, sketch the curve in each of the following cases.
   (A) \(m = 1 , n = 1\),
   (B) \(m = 2 , n = 2\),
   (C) \(m = 2 , n = 4\),
   (D) \(m = 4 , n = 2\).
2. What feature does the curve have when \(m = n\) ? What is the effect on the curve of interchanging \(m\) and \(n\) when \(m \neq n\) ?
3. Describe how the \(x\)-coordinate of the maximum on the curve varies as \(m\) and \(n\) vary. Use calculus to determine the \(x\)-coordinate of the maximum.
4. Find the condition on \(m\) for the gradient to be zero when \(x = 0\). State a corresponding result for the gradient to be zero when \(x = 1\).
5. Use your calculator to investigate the shape of the curve for large values of \(m\) and \(n\). Hence conjecture what happens to the value of the integral \(\int _ { 0 } ^ { 1 } x ^ { m } ( 1 - x ) ^ { n } \mathrm {~d} x\) as \(m\) and \(n\) tend to infinity.
6. Use your calculator to investigate the shape of the curve for small values of \(m\) and \(n\). Hence conjecture what happens to the shape of the curve as \(m\) and \(n\) tend to zero. }{www.ocr.org.uk}) after the live examination series.
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6

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

9 The positive constant \(a\) is such that \(\int _ { a } ^ { 2 a } \frac { 2 x ^ { 3 } - 5 x ^ { 2 } + 4 } { x ^ { 2 } } \mathrm {~d} x = 0\).

1. Show that \(3 a ^ { 3 } - 5 a ^ { 2 } + 2 = 0\).
2. Show that \(a = 1\) is a root of \(3 a ^ { 3 } - 5 a ^ { 2 } + 2 = 0\), and hence find the other possible value of \(a\), giving your answer in simplified surd form.

6

1. Use integration to find the exact area of the region enclosed by the curve \(y = x ^ { 2 } + 4 x\), the \(x\)-axis and the lines \(x = 3\) and \(x = 5\).
2. Find \(\int ( 2 - 6 \sqrt { y } ) \mathrm { d } y\).
3. Evaluate \(\int _ { 1 } ^ { \infty } \frac { 8 } { x ^ { 3 } } \mathrm {~d} x\).

6

1. Use the trapezium rule, with 2 strips each of width 4 , to show that an approximate value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\) is \(32 + 16 \sqrt { 5 }\).
2. Use a sketch graph to explain why the actual value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\) is greater than \(32 + 16 \sqrt { 5 }\).
3. Use integration to find the exact value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\).

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 \(\int _ { 1 } ^ { \infty } \left( 6 x ^ { - 2 } - 4 x ^ { - 3 } \right) \mathrm { d } x\).

10 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h] \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.

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

5 \includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-02_559_1191_1749_479} The diagram shows the curve with equation \(y = \frac { 6 } { \sqrt { 3 x - 2 } }\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 1 , x = a\) and \(y = 0\), where \(a\) is a constant greater than 1 . It is given that the area of \(R\) is 16 square units. Find the value of \(a\) and hence find the exact volume of the solid formed when \(R\) is rotated completely about the \(x\)-axis.
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1 Show that \(\int _ { \sqrt { 2 } } ^ { \sqrt { 6 } } \frac { 2 } { x } \mathrm {~d} x = \ln 3\).

4

1. Show that \(\int _ { 0 } ^ { 4 } \frac { 18 } { \sqrt { 6 x + 1 } } \mathrm {~d} x = 24\).
2. Find \(\int _ { 0 } ^ { 1 } \left( \mathrm { e } ^ { x } + 2 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in terms of e .

7 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_547_851_1749_605} The diagram shows the curve \(y = \sqrt { \frac { 3 } { 4 x + 1 } }\) for \(0 \leqslant x \leqslant 20\). The point \(P\) on the curve has coordinates \(\left( 20 , \frac { 1 } { 9 } \sqrt { 3 } \right)\). The shaded region \(R\) is enclosed by the curve and the lines \(x = 0\) and \(y = \frac { 1 } { 9 } \sqrt { 3 }\).

1. Find the exact area of \(R\).
2. Find the exact volume of the solid obtained when \(R\) is rotated completely about the \(x\)-axis.

9 Fig. 9 shows the curve \(y = \frac { x ^ { 2 } } { 3 x - 1 }\).
P is a turning point, and the curve has a vertical asymptote \(x = a\). \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }\).
3. Find the exact coordinates of the turning point P . Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point.
4. Using the substitution \(u = 3 x - 1\), show that \(\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 1\).

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 \(x = 1\).

4 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( 1 + \sin x ) ^ { 2 } \mathrm {~d} x\).

7 A particle \(P\) is projected from a fixed point \(O\) on a straight line. The displacement \(x\) m of \(P\) from \(O\) at time \(t \mathrm {~s}\) after projection is given by \(x = 0.1 t ^ { 3 } - 0.3 t ^ { 2 } + 0.2 t\).

1. Express the velocity and acceleration of \(P\) in terms of \(t\).
2. Show that when the acceleration of \(P\) is zero, \(P\) is at \(O\).
3. Find the values of \(t\) when \(P\) is stationary. At the instant when \(P\) first leaves \(O\), a particle \(Q\) is projected from \(O\). \(Q\) moves on the same straight line as \(P\) and at time \(t \mathrm {~s}\) after projection the velocity of \(Q\) is given by \(\left( 0.2 t ^ { 2 } - 0.4 \right) \mathrm { ms } ^ { - 1 } . P\) and \(Q\) collide first when \(t = T\).
4. Show that \(T\) satisfies the equation \(t ^ { 2 } - 9 t + 18 = 0\), and hence find \(T\).

7 \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-4_782_1065_251_500} The diagram shows the curve \(y = \frac { 1 } { x }\) for \(x > 0\) and a set of \(( n - 1 )\) rectangles of unit width below the curve. These rectangles can be used to obtain an inequality of the form $$\frac { 1 } { a } + \frac { 1 } { a + 1 } + \frac { 1 } { a + 2 } + \ldots + \frac { 1 } { b } < \int _ { 1 } ^ { n } \frac { 1 } { x } \mathrm {~d} x$$ Another set of rectangles can be used similarly to obtain $$\int _ { 1 } ^ { n } \frac { 1 } { x } \mathrm {~d} x < \frac { 1 } { c } + \frac { 1 } { c + 1 } + \frac { 1 } { c + 2 } + \ldots + \frac { 1 } { d }$$

1. Write down the values of the constants \(a\) and \(c\), and express \(b\) and \(d\) in terms of \(n\). The function f is defined by \(\mathrm { f } ( n ) = 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \ldots + \frac { 1 } { n } - \ln n\), for positive integers \(n\).
2. Use your answers to part (i) to obtain upper and lower bounds for \(\mathrm { f } ( n )\).
3. By using the first 2 terms of the Maclaurin series for \(\ln ( 1 + x )\) show that, for large \(n\), $$f ( n + 1 ) - f ( n ) \approx - \frac { n - 1 } { 2 n ^ { 2 } ( n + 1 ) } .$$

6 \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-4_656_1017_251_525} The diagram shows part of the curve \(y = \ln ( \ln ( x ) )\). The region between the curve and the \(x\)-axis for \(3 \leqslant x \leqslant 6\) is shaded.

1. By considering \(n\) rectangles of equal width, show that a lower bound, \(L\), for the area of the shaded region is \(\frac { 3 } { n } \sum _ { r = 0 } ^ { n - 1 } \ln \left( \ln \left( 3 + \frac { 3 r } { n } \right) \right)\).
2. By considering another set of \(n\) rectangles of equal width, find a similar expression for an upper bound, \(U\), for the area of the shaded region.
3. Find the least value of \(n\) for which \(U - L < 0.001\).

7 By considering the binomial expansion of \(\left( z - \frac { 1 } { z } \right) ^ { 6 }\), where \(z = \cos \theta + \mathrm { i } \sin \theta\), express \(\sin ^ { 6 } \theta\) in the form $$\frac { 1 } { 32 } ( p + q \cos 2 \theta + r \cos 4 \theta + s \cos 6 \theta ) ,$$ where \(p , q , r\) and \(s\) are integers to be determined. Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } \theta \mathrm {~d} \theta\).

6 Let \(I _ { n }\) denote \(\int _ { 0 } ^ { 2 } \left( 4 + x ^ { 2 } \right) ^ { - n } \mathrm {~d} x\).

1. Find \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x \left( 4 + x ^ { 2 } \right) ^ { - n } \right)\) and hence show that $$8 n I _ { n + 1 } = ( 2 n - 1 ) I _ { n } + 2 \times 8 ^ { - n } .$$
2. Use the result for integrating \(\frac { 1 } { x ^ { 2 } + a ^ { 2 } }\) with respect to \(x\), in the List of Formulae (MF10), to find the value of \(I _ { 1 }\) and deduce that $$I _ { 3 } = \frac { 3 } { 1024 } \pi + \frac { 1 } { 128 }$$

9 It is given that $$I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + x ^ { 3 } \right) ^ { - n } \mathrm {~d} x$$ where \(n > 0\).

1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ x \left( 1 + x ^ { 3 } \right) ^ { - n } \right] = - ( 3 n - 1 ) \left( 1 + x ^ { 3 } \right) ^ { - n } + 3 n \left( 1 + x ^ { 3 } \right) ^ { - n - 1 }$$ and hence, or otherwise, show that $$I _ { n + 1 } = \frac { 2 ^ { - n } } { 3 n } + \left( 1 - \frac { 1 } { 3 n } \right) I _ { n }$$
2. By considering the graph of \(y = \frac { 1 } { 1 + x ^ { 3 } }\), show that \(I _ { 1 } < 1\).
3. Deduce that \(I _ { 3 } < \frac { 53 } { 72 }\).

7 Use de Moivre's theorem to express \(\sin ^ { 6 } \theta\) in the form $$a + b \cos 2 \theta + c \cos 4 \theta + d \cos 6 \theta$$ where \(a , b , c , d\) are constants to be found. Hence evaluate $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } 2 x d x$$ leaving your answer in terms of \(\pi\).

The curve \(C\) has equation \(y = \frac { 1 } { 3 } x ^ { \frac { 1 } { 2 } } ( 3 - x )\), for \(0 \leqslant x \leqslant 3\). Find the mean value of \(y\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant 3\). Show that $$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \left( x ^ { - \frac { 1 } { 2 } } + x ^ { \frac { 1 } { 2 } } \right)$$ where \(s\) denotes arc length, and find the arc length of \(C\). Find the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

2 The curve \(C\) has equation \(y = 2 x ^ { \frac { 1 } { 2 } }\) for \(0 \leqslant x \leqslant 4\). Find

1. the mean value of \(y\) with respect to \(x\) for \(0 \leqslant x \leqslant 4\),
2. the \(y\)-coordinate of the centroid of the region enclosed by \(C\), the line \(x = 4\) and the \(x\)-axis.

Let \(I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + x ^ { 2 } \right) ^ { n } \mathrm {~d} x\). Show that, for all integers \(n\), $$( 2 n + 1 ) I _ { n } = 2 n I _ { n - 1 } + 2 ^ { n }$$ Evaluate \(I _ { 0 }\) and hence find \(I _ { 3 }\). Given that \(I _ { - 1 } = \frac { 1 } { 4 } \pi\), find \(I _ { - 3 }\).