1.08d Evaluate definite integrals: between limits

6 The graph of the probability density function f of a random variable \(X\) is symmetrical about the line \(x = 2\). It is given that \(\mathrm { P } ( 2 < X < 5 ) = \frac { 117 } { 256 }\).

1. Using only this information show that \(\mathrm { P } ( X > - 1 ) = \frac { 245 } { 256 }\).
   It is now given that, for \(x\) in a suitable domain, $$f ( x ) = k \left( 12 + 4 x - x ^ { 2 } \right) , \text { where } k \text { is a constant. }$$
2. Find the value of \(k\).
3. A different random variable \(X\) has probability density function \(\mathbf { g } ( x ) = \frac { 2 } { 9 } \left( 2 + x - x ^ { 2 } \right)\). The domain of \(X\) is all values of \(x\) for which \(\mathrm { g } ( x ) \geqslant 0\). Find \(\operatorname { Var } ( X )\). \includegraphics[max width=\textwidth, alt={}, center]{ff3433b0-baab-45e3-845e-56a794739bba-11_63_1547_447_347}

3. Find, using calculus and showing each step of your working, $$\int _ { 1 } ^ { 4 } \left( 6 x - 3 - \frac { 2 } { \sqrt { x } } \right) \mathrm { d } x$$

7. (i) Find $$\int \frac { 2 + 4 x ^ { 3 } } { x ^ { 2 } } \mathrm {~d} x$$ giving each term in its simplest form.
(ii) Given that \(k\) is a constant and $$\int _ { 2 } ^ { 4 } \left( \frac { 4 } { \sqrt { x } } + k \right) \mathrm { d } x = 30$$ find the exact value of \(k\).

8. Given \(k > 3\) and $$\int _ { 3 } ^ { k } \left( 2 x + \frac { 6 } { x ^ { 2 } } \right) \mathrm { d } x = 10 k$$ show that \(k ^ { 3 } - 10 k ^ { 2 } - 7 k - 6 = 0\)

7. (i) A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(( 2,3 )\). Given that $$f ^ { \prime } ( x ) = \frac { 4 } { x ^ { 3 } } + 2 x - 1$$ find the value of \(\mathrm { f } ( 1 )\).
(ii) Given that $$\int _ { 1 } ^ { 4 } ( 3 \sqrt { x } + A ) \mathrm { d } x = 21$$ find the exact value of the constant \(A\).

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { x + 2 } , x \geqslant - 2\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line \(x = 6\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { x + 2 }\)

1. Complete the table above, giving the missing value of \(y\) to 4 decimal places.
2. Use the trapezium rule, with all of the values of \(y\) in the completed table, to find an approximate value for the area of \(R\), giving your answer to 3 decimal places. Use your answer to part (b) to find approximate values of
3. 1. \(\int _ { - 2 } ^ { 6 } \frac { \sqrt { x + 2 } } { 2 } \mathrm {~d} x\)
   2. \(\int _ { - 2 } ^ { 6 } ( 2 + \sqrt { x + 2 } ) \mathrm { d } x\)

8. (a) Find \(\int \left( 3 x ^ { 2 } + 4 x - 15 \right) \mathrm { d } x\), simplifying each term. Given that \(b\) is a constant and $$\int _ { b } ^ { 4 } \left( 3 x ^ { 2 } + 4 x - 15 \right) \mathrm { d } x = 36$$ (b) show that \(b ^ { 3 } + 2 b ^ { 2 } - 15 b = 0\) (c) Hence find the possible values of \(b\).

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

1. The curve \(C\) has equation

$$y = \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \quad x > 0$$ where \(k\) is a positive constant.

1. Show that $$\int _ { 1 } ^ { 16 } \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \mathrm {~d} x = a k ^ { 2 } + b k + \frac { 2046 } { 5 }$$ where \(a\) and \(b\) are integers to be found. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0e3b364c-151b-471d-acb6-01afb018fb75-26_645_670_904_699} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of the curve \(C\) and the line \(l\).
   Given that \(l\) intersects \(C\) at the point \(A ( 1,9 )\) and at the point \(B ( 16 , q )\) where \(q\) is a constant,
2. show that \(k = 4\) The region \(R\), shown shaded in Figure 1, is bounded by \(C\) and \(l\) Using the answers to parts (a) and (b),
3. find the area of region \(R\)

9. \begin{figure}[h] \end{figure} Figure 1 is a sketch of the curve \(C\) with equation $$y = 2 x ^ { \frac { 3 } { 2 } } ( 4 - x ) \quad x \geqslant 0$$ The point \(P\) is the stationary point of \(C\).

1. Find, using calculus, the \(x\) coordinate of \(P\). The region \(R _ { 1 }\), shown shaded in Figure 1, is bounded by \(C\) and the \(x\)-axis.
   The region \(R _ { 2 }\), also shown shaded in Figure 1, is bounded by \(C\), the \(x\)-axis and the line with equation \(x = k\), where \(k\) is a constant. Given that the area of \(R _ { 1 }\) is equal to the area of \(R _ { 2 }\)
2. find, using calculus, the exact value of \(k\).

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

1. $$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 5$$ Find

1. \(\mathrm { f } ^ { \prime \prime } ( x )\),
2. \(\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x\).

9. \(y\) \begin{figure}[h] \end{figure} The finite region \(R\), as shown in Figure 2, is bounded by the \(x\)-axis and the curve with equation $$y = 27 - 2 x - 9 \sqrt { } x - \frac { 16 } { x ^ { 2 } } , \quad x > 0$$ The curve crosses the \(x\)-axis at the points \(( 1,0 )\) and \(( 4,0 )\).

1. Complete the table below, by giving your values of \(y\) to 3 decimal places.

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	0	5.866		5.210		1.856	0

2. Use the trapezium rule with all the values in the completed table to find an approximate value for the area of \(R\), giving your answer to 2 decimal places.
3. Use integration to find the exact value for the area of \(R\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x ^ { 3 } - 6 x ^ { 2 } + 9 x + 5$$ The point \(P ( 4,9 )\) lies on \(C\).

1. Show that the normal to \(C\) at the point \(P\) has equation $$x + 9 y = 85$$ The region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the \(y\)-axis and the normal to \(C\) at \(P\).
2. Showing all your working, calculate the exact area of \(R\).

10. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve \(C\) with equation \(y = 2 x + \frac { 8 } { x ^ { 2 } } - 5 , x > 0\).
The points \(P\) and \(Q\) lie on \(C\) and have \(x\)-coordinates 1 and 4 respectively. The region \(R\), shaded in Figure 1, is bounded by \(C\) and the straight line joining \(P\) and \(Q\).

1. Find the exact area of \(R\).
2. Use calculus to show that \(y\) is increasing for \(x > 2\).

Evaluate \(\int _ { 1 } ^ { 8 } \frac { 1 } { \sqrt { } x } \mathrm {~d} x\), giving your answer in the form \(a + b \sqrt { } 2\), where \(a\) and \(b\) are integers.

1. Use calculus to find the value of

$$\int _ { 1 } ^ { 4 } ( 2 x + 3 \sqrt { } x ) d x$$

1. Use integration to find

$$\int _ { 1 } ^ { \sqrt { 3 } } \left( \frac { x ^ { 3 } } { 6 } + \frac { 1 } { 3 x ^ { 2 } } \right) \mathrm { d } x$$ giving your answer in the form \(a + b \sqrt { 3 }\), where \(a\) and \(b\) are constants to be determined.

1. (i) Find

$$\int \frac { 12 } { ( 2 x - 1 ) ^ { 2 } } \mathrm {~d} x$$ giving your answer in simplest form.
(ii) (a) Write \(\frac { 4 x + 3 } { x + 2 }\) in the form $$A + \frac { B } { x + 2 } \text { where } A \text { and } B \text { are constants to be found }$$ (b) Hence find, using algebraic integration, the exact value of $$\int _ { - 8 } ^ { - 5 } \frac { 4 x + 3 } { x + 2 } d x$$ giving your answer in simplest form.

3. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Given that \(k\) is a positive constant,

1. find $$\int \frac { 9 x } { 3 x ^ { 2 } + k } d x$$ Given also that $$\int _ { 2 } ^ { 5 } \frac { 9 x } { 3 x ^ { 2 } + k } \mathrm {~d} x = \ln 8$$
2. find the value of \(k\)

8. \begin{figure}[h] \end{figure} Figure 4 is a graph showing the velocity of a sprinter during a 100 m race.
The sprinter's velocity during the race, \(v \mathrm {~ms} ^ { - 1 }\), is modelled by the equation $$v = 12 - \mathrm { e } ^ { t - 10 } - 12 \mathrm { e } ^ { - 0.75 t } \quad t \geqslant 0$$ where \(t\) seconds is the time after the sprinter begins to run. According to the model,

1. find, using calculus, the sprinter's maximum velocity during the race. Given that the sprinter runs 100 m in \(T\) seconds, such that $$\int _ { 0 } ^ { T } v \mathrm {~d} t = 100$$
2. show that \(T\) is a solution of the equation $$T = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T } + \mathrm { e } ^ { T - 10 } - \mathrm { e } ^ { - 10 } \right)$$ The iteration formula $$T _ { n + 1 } = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T _ { n } } + \mathrm { e } ^ { T _ { n } - 10 } - \mathrm { e } ^ { - 10 } \right)$$ is used to find an approximate value for \(T\) Using this iteration formula with \(T _ { 1 } = 10\)
3. find, to 4 decimal places,
   1. the value of \(T _ { 2 }\)
   2. the time taken by the sprinter to run the race, according to the model.

1. (i) Find \(\frac { \mathrm { d } } { \mathrm { d } x } \ln \left( \sin ^ { 2 } 3 x \right)\) writing your answer in simplest form.
   (ii) (a) Find \(\frac { \mathrm { d } } { \mathrm { d } x } \left( 3 x ^ { 2 } - 4 \right) ^ { 6 }\) (b) Hence show that

$$\int _ { 0 } ^ { \sqrt { 2 } } x \left( 3 x ^ { 2 } - 4 \right) ^ { 5 } \mathrm {~d} x = R$$ where \(R\) is an integer to be found.
(Solutions relying on calculator technology are not acceptable.)

2. $$g ( x ) = \frac { 2 x ^ { 2 } - 5 x + 8 } { x - 2 }$$

1. Write \(g ( x )\) in the form $$A x + B + \frac { C } { x - 2 }$$ where \(A , B\) and \(C\) are integers to be found.
2. Hence use algebraic integration to show that $$\int _ { 4 } ^ { 8 } \mathrm {~g} ( x ) \mathrm { d } x = \alpha + \beta \ln 3$$ where \(\alpha\) and \(\beta\) are integers to be found.

9. (a) Given that $$\frac { x ^ { 4 } - x ^ { 3 } - 10 x ^ { 2 } + 3 x - 9 } { x ^ { 2 } - x - 12 } \equiv x ^ { 2 } + P + \frac { Q } { x - 4 } \quad x > - 3$$ find the value of the constant \(P\) and show that \(Q = 5\) The curve \(C\) has equation \(y = \mathrm { g } ( x )\), where $$g ( x ) = \frac { x ^ { 4 } - x ^ { 3 } - 10 x ^ { 2 } + 3 x - 9 } { x ^ { 2 } - x - 12 } \quad - 3 < x < 3.5 \quad x \in \mathbb { R }$$ (b) Find the equation of the tangent to \(C\) at the point where \(x = 2\) Give your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants to be found. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\).
The region \(R\), shown shaded in Figure 4, is bounded by \(C\), the \(y\)-axis, the \(x\)-axis and the line with equation \(x = 2\) (c) Find the exact area of \(R\), writing your answer in the form \(a + b \ln 2\), where \(a\) and \(b\) are constants to be found. \includegraphics[max width=\textwidth, alt={}, center]{96948fd3-5438-4e95-b41b-2f649ca8dfac-31_2255_50_314_34}
\includegraphics[max width=\textwidth, alt={}, center]{96948fd3-5438-4e95-b41b-2f649ca8dfac-32_106_113_2524_1832}

3. \begin{figure}[h] \end{figure} The total mass of gold, \(G\) tonnes, extracted from a mine is modelled by the equation $$G = 40 - 30 \mathrm { e } ^ { 1 - 0.05 t } \quad t \geqslant k \quad G \geqslant 0$$ where \(t\) is the number of years after 1st January 1800.
Figure 2 shows a sketch of \(G\) against \(t\). Use the equation of the model to answer parts (a), (b) and (c).

1. 1. Find the value of \(k\).
   2. Hence find the year and month in which gold started being extracted from the mine.
2. Find the total mass of gold extracted from the mine up to 1st January 1870. There is a limit to the mass of gold that can be extracted from the mine.
3. State the value of this limit.
   M