1.08d Evaluate definite integrals: between limits

4 Evaluate the following integrals, giving your answers in exact form. \begin{displayquote}

1. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x ^ { 2 } + 1 } \mathrm {~d} x\)
2. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x + 1 } \mathrm {~d} x\) \end{displayquote}

8

1. Evaluate \(A _ { 0 } = \int _ { 0 } ^ { 2 } \left( 2 + 2 x - x ^ { 2 } \right) \mathrm { d } x\). Fig 8.1 illustrates the cross-section of a proposed tunnel. Lengths are in metres. The equation of the curved section is \(y = 2 + \sqrt { 2 x - x ^ { 2 } }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_419_515_1155_836} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
   \end{figure} The designers need to know the area of the cross-section, \(A \mathrm {~m} ^ { 2 }\), so that they can work out the volume of the soil that will need to be removed when the tunnel is built.
2. An initial estimate, \(A _ { 1 }\), is given by the area of the 8 rectangles shown in Fig 8.2. Calculate \(A _ { 1 }\), and state whether it is an overestimate or underestimate for \(A\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_520_645_2053_644} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure}
3. On graph paper, draw the graphs of $$y = 2 + 2 x - x ^ { 2 } \text { and } y = 2 + \sqrt { 2 x - x ^ { 2 } } \text { for } 0 \leq x \leq 2 .$$ Make it clear which equation applies to which curve.
4. State whether \(A _ { 0 }\), your answer to part (i), is an underestimate for \(A\) or an overestimate. Give a reason for your answer.
5. The designers use the trapezium rule to estimate \(A\). What values does this give when they take
   (A) 2 strips,
   (B) 4 strips,
   (C) 8 strips? What can you conclude about the value of \(A\) ?
6. The best estimate from the trapezium rule is denoted by \(A _ { 2 }\). State, with a reason, whether the true value of \(A\) is nearer \(A _ { 1 }\) or \(A _ { 2 }\).

8 Scientists predict the velocity ( \(v\) kilometres per minute) for the new "outer explorer" spacecraft over the first minute of its entry to the atmosphere of the planet Titan to be modelled by the equation: $$v = \frac { 5000 } { ( 1 + t ) ( 2 + t ) ^ { 2 } } , 0 \leq t \leq 1 \text { where } t \text { represents time in minutes. }$$

1. Use a binomial expansion to expand \(( 1 + t ) ^ { - 1 }\) up to and including the term in \(t ^ { 2 }\).
2. Use a binomial expansion to expand \(( 2 + t ) ^ { - 2 }\) up to and including the term in \(t ^ { 2 }\).
3. Hence, or otherwise, show that \(v \approx 1250 \left( 1 - 2 t + \frac { 11 t ^ { 2 } } { 4 } \right)\).
4. The displacement of the spacecraft can be found by calculating the area under the velocity time graph. Use the approximation found in part (iii) to estimate the displacement of the spacecraft over the first half minute.
5. Write \(\frac { 1 } { ( 1 + t ) ( 2 + t ) ^ { 2 } }\) in partial fractions.
6. The displacement of the spacecraft in the first \(T\) minutes is given by \(\int _ { 0 } ^ { T } v \mathrm {~d} t\) Calculate the exact value of the displacement of the spacecraft over the first half minute given by the model.
7. On further investigation the scientists believe the original model may be valid for up to three minutes. Explain why the approximation in (iii) will be no longer be valid for this time interval.

1. Evaluate

$$\int _ { 0 } ^ { \pi } \sin x ( 1 + \cos x ) d x$$

9. $$f ( x ) = \frac { 8 - x } { ( 1 + x ) ( 2 - x ) } , \quad | x | < 1$$

1. Express \(\mathrm { f } ( x )\) in partial fractions.
2. Show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x = \ln k$$ where \(k\) is an integer to be found.
3. Find the series expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\), simplifying each coefficient.

4.

1. Express \(\frac { 3 x + 6 } { 3 x - x ^ { 2 } }\) in partial fractions.
2. Evaluate \(\int _ { 1 } ^ { 2 } \frac { 3 x + 6 } { 3 x - x ^ { 2 } } \mathrm {~d} x\).

3. Using the substitution \(u = \mathrm { e } ^ { x } - 1\), show that $$\int _ { \ln 2 } ^ { \ln 5 } \frac { \mathrm { e } ^ { 2 x } } { \sqrt { \mathrm { e } ^ { x } - 1 } } \mathrm {~d} x = \frac { 20 } { 3 }$$

5. $$f ( x ) = \frac { 7 + 3 x + 2 x ^ { 2 } } { ( 1 - 2 x ) ( 1 + x ) ^ { 2 } } , \quad | x | > \frac { 1 } { 2 }$$

1. Express \(\mathrm { f } ( x )\) in partial fractions.
2. Show that $$\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x = p - \ln q$$ where \(p\) is rational and \(q\) is an integer.

7

1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 - x ^ { 2 } } }\) for \(| x | < 2\). [4]
2. Use this result to find an approximation for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to
   4 significant figures.
3. Given that \(\int \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c\), evaluate \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to 4 significant figures.

10 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.

1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.

7 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

4 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

4

1. Given that $$y = x \sqrt { 1 - x ^ { 2 } } - \cos ^ { - 1 } x$$ find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in a simplified form.
2. Hence, or otherwise, find the exact value of \(\int _ { 0 } ^ { 1 } 2 \sqrt { 1 - x ^ { 2 } } \mathrm {~d} x\).

8

1. Use the substitution \(t = \tan \frac { 1 } { 2 } x\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x = 2 \sqrt { } 2 \int _ { 0 } ^ { 1 } \frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) } \mathrm { d } t$$
2. Express \(\frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) }\) in partial fractions.
3. Hence find \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x\), expressing your answer in an exact form.

1

1. A curve has polar equation \(r = a \mathrm { e } ^ { - k \theta }\) for \(0 \leqslant \theta \leqslant \pi\), where \(a\) and \(k\) are positive constants. The points A and B on the curve correspond to \(\theta = 0\) and \(\theta = \pi\) respectively.
   1. Sketch the curve.
   2. Find the area of the region enclosed by the curve and the line AB .
2. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { 3 + 4 x ^ { 2 } } \mathrm {~d} x\).
3. 1. Find the Maclaurin series for \(\tan x\), up to the term in \(x ^ { 3 }\).
   2. Use this Maclaurin series to show that, when \(h\) is small, \(\int _ { h } ^ { 4 h } \frac { \tan x } { x } \mathrm {~d} x \approx 3 h + 7 h ^ { 3 }\).

2.(a)On the same diagram,sketch \(y = x\) and \(y = \sqrt { } x\) ,for \(x \geq 0\) ,and mark clearly the coordinates of the points of intersection of the two graphs.
(b)With reference to your sketch,explain why there exists a value \(a\) of \(x ( a > 1 )\) such that $$\int _ { 0 } ^ { a } x \mathrm {~d} x = \int _ { 0 } ^ { a } \sqrt { } x \mathrm {~d} x$$ (c)Find the exact value of \(a\) .
(d)Hence,or otherwise,find a non-constant function \(\mathrm { f } ( x )\) and a constant \(b ( b \neq 0 )\) such that $$\int _ { - b } ^ { b } \mathrm { f } ( x ) \mathrm { d } x = \int _ { - b } ^ { b } \sqrt { } [ \mathrm { f } ( x ) ] \mathrm { d } x$$

7. $$\mathrm { f } ( x ) = \left[ 1 + \cos \left( x + \frac { \pi } { 4 } \right) \right] \left[ 1 + \sin \left( x + \frac { \pi } { 4 } \right) \right] , \quad 0 \leqslant x \leqslant 2 \pi$$

1. Show that \(\mathrm { f } ( x )\) may be written in the form $$f ( x ) = \left( \frac { 1 } { \sqrt { 2 } } + \cos x \right) ^ { 2 } , \quad 0 \leqslant x \leqslant 2 \pi$$
2. Find the range of the function \(\mathrm { f } ( x )\). The graph of \(y = \mathrm { f } ( x )\) is shown in Figure 2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0396f61a-b844-40ed-98d1-82ee2d8a6807-5_426_938_849_591} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure}
3. Find the coordinates of all the maximum and minimum points on this curve. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0396f61a-b844-40ed-98d1-82ee2d8a6807-5_432_942_1535_589} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} The region \(R\), bounded by \(y = 2\) and \(y = \mathrm { f } ( x )\), is shown shaded in Figure 3.
4. Find the area of \(R\).

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 $$\frac { 144 } { 55 } \leqslant \int _ { - 1 } ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$$

6. \includegraphics[max width=\textwidth, alt={}, center]{0214eebf-93f2-4338-9222-443000115225-4_346_1040_303_548} \section*{Figure 1} Figure 1 shows a sketch of the curve \(C _ { 1 }\) with equation $$y = \cos ( \cos x ) \sin x \quad \text { for } \quad 0 \leqslant x \leqslant \pi$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Hence verify that the turning point is at \(x = \frac { \pi } { 2 }\) and find the \(y\) coordinate of this point.
3. Find the area of the region bounded by \(C _ { 1 }\) and the positive \(x\)-axis between \(x = 0\) and \(x = \pi\) Figure 2 shows a sketch of the curve \(C _ { 1 }\) and the curve \(C _ { 2 }\) with equation $$y = \sin ( \cos x ) \sin x \quad \text { for } \quad 0 \leqslant x \leqslant \pi$$ \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0214eebf-93f2-4338-9222-443000115225-4_519_1065_1631_484} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the origin and the point \(A ( a , b )\) ,where \(a < \pi\)
4. Find \(a\) and \(b\) ,giving \(b\) in a form not involving trigonometric functions.
5. Find the area of the shaded region between \(C _ { 1 }\) and \(C _ { 2 }\)

7. \begin{figure}[h] \end{figure} Diagram not drawn to scale Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 1 } { \sqrt { 2 x + 5 } } , x > - 2.5\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 5\)

1. Use the trapezium rule with three strips of equal width to find an estimate for the area of \(R\), giving your answer to 3 decimal places.
2. Use calculus to find the exact area of \(R\).
3. Hence calculate the magnitude of the error of the estimate found in part (a), giving your answer to one significant figure.

9 \includegraphics[max width=\textwidth, alt={}, center]{c1eee696-3d7f-410a-91a8-fa902309c117-16_307_593_269_735} The diagram shows the curve \(y = \sin ^ { 2 } 2 x \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \sin x\), find the area of the shaded region bounded by the curve and the \(x\)-axis.
   

5. \begin{figure}[h] \end{figure} Show that the area of the finite region between the curves \(y = \tan ^ { 2 } x\) and \(y = 4 \cos 2 x - 1\) in the interval \(- \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\), shown shaded in Figure 3, is given by $$2 \sqrt { 2 \sqrt { 3 } } - 2 \sqrt { 2 \sqrt { 3 } - 3 }$$

7. \begin{figure}[h] \end{figure} A circular tower of radius 1 metre stands in a large horizontal field of grass.A goat is attached to one end of a rope and the other end of the rope is attached to a fixed point \(O\) at the base of the tower.The goat cannot enter the tower. Taking the point \(O\) as the origin( 0,0 ),the centre of the base of the tower is at the point \(T ( 0,1 )\) ,where the unit of length is the metre. The rope has length \(\pi\) metres and you may ignore the size of the goat.
The curve \(C\) shown in Figure 4 represents the edge of the region that the goat can reach.

1. Write down the equation of \(C\) for \(y < 0\) When the goat is at the point \(G ( x , y )\) ,with \(x > 0\) and \(y > 0\) ,as shown in Figure 4 ,the rope lies along \(O A G\) where \(O A\) is an arc of the circle with angle \(O T A = \theta\) radians and \(A G\) is a tangent to the circle at \(A\) .
2. With the aid of a suitable diagram show that $$\begin{aligned} & x = \sin \theta + ( \pi - \theta ) \cos \theta \\ & y = 1 - \cos \theta + ( \pi - \theta ) \sin \theta \end{aligned}$$
3. By considering \(\int y \frac { \mathrm {~d} x } { \mathrm {~d} \theta } \mathrm {~d} \theta\), show that the area, in the first quadrant, between \(C\), the positive \(x\)-axis and the positive \(y\)-axis can be expressed in the form $$\int _ { 0 } ^ { \pi } u \sin u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u$$
4. Show that \(\int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u = \frac { \pi ^ { 3 } } { 6 } + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u\)
5. Hence find the area of grass that can be reached by the goat.

8

1. By expressing \(\sin \theta\) in terms of \(\mathrm { e } ^ { \mathrm { i } \theta }\) and \(\mathrm { e } ^ { - \mathrm { i } \theta }\), show that $$\sin ^ { 6 } \theta \equiv - \frac { 1 } { 32 } ( \cos 6 \theta - 6 \cos 4 \theta + 15 \cos 2 \theta - 10 )$$
2. Replace \(\theta\) by ( \(\frac { 1 } { 2 } \pi - \theta\) ) in the identity in part (i) to obtain a similar identity for \(\cos ^ { 6 } \theta\).
3. Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \sin ^ { 6 } \theta - \cos ^ { 6 } \theta \right) \mathrm { d } \theta\).