1.08d Evaluate definite integrals: between limits

6 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-3_586_798_267_676} The diagram shows the curves \(y = \mathrm { e } ^ { 3 x }\) and \(y = ( 2 x - 1 ) ^ { 4 }\). The shaded region is bounded by the two curves and the line \(x = \frac { 1 } { 2 }\). The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced.

4 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-2_419_707_1576_660} The diagram shows the curve $$y = \frac { 1 } { \sqrt { } ( 4 x + 1 ) }$$ The region \(R\) (shaded in the diagram) is enclosed by the curve, the axes and the line \(x = 2\).

1. Show that the exact area of \(R\) is 1 .
2. The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed.

7 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE . \section*{[Question 8 is printed overleaf.]}

7 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

8 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \ln ( 1 + \sqrt { x } )\) for \(x \geqslant 0\).
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).

6.

1. Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius \(r\) is at a distance \(\frac { 3 } { 8 } r\) from the centre of its plane face.
   [0pt] [You may assume that the volume of a sphere of radius \(r\) is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ]
   (5) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ffe0bc72-3136-48d9-9d5b-4a364d134070-09_351_597_598_678} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} A uniform solid hemisphere of mass \(m\) and radius \(r\) is joined to a uniform solid right circular cone to form a solid \(S\). The cone has mass \(M\), base radius \(r\) and height \(4 r\). The vertex of the cone is \(O\). The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.
2. Find the distance of the centre of mass of \(S\) from \(O\). The point \(A\) lies on the circumference of the base of the cone. The solid is placed on a horizontal table with \(O A\) in contact with the table. The solid remains in equilibrium in this position.
3. Show that \(M \geqslant \frac { 1 } { 10 } m\)

1. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y ^ { 2 } = 9 ( 4 - x )\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 1. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.
Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).

5. \begin{figure}[h] \end{figure} Figure 3 shows the finite region \(R\) which is bounded by part of the curve with equation \(y = \sin x\), the \(x\)-axis and the line with equation \(x = \frac { \pi } { 2 }\). A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. find, in terms of \(\pi\), the \(x\) coordinate of the centre of mass of \(S\).

5. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 3, is bounded by the circle with centre \(O\) and radius \(r\), the line with equation \(x = \frac { 3 } { 5 } r\) and the \(x\)-axis. The region is rotated through one complete revolution about the \(x\)-axis to form a uniform solid \(S\).

1. Use algebraic integration to show that the \(x\) coordinate of the centre of mass of \(S\) is \(\frac { 48 } { 65 } r\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ae189c40-0071-4a6b-91eb-8ffebe082a04-16_394_643_1311_653} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} A bowl is made from a uniform solid hemisphere of radius 6 cm by removing a hemisphere of radius 5 cm . Both hemispheres have the same centre \(A\) and the same axis of symmetry. The bowl is fixed with its open plane face uppermost and horizontal. Liquid is poured into the bowl. The depth of the liquid is 2 cm , as shown in Figure 4. The mass of the empty bowl is \(5 M \mathrm {~kg}\) and the mass of the liquid is \(2 M \mathrm {~kg}\).
2. Find, to 3 significant figures, the distance from \(A\) to the centre of mass of the bowl with its liquid.

1. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by the curve with equation \(y = \frac { 1 } { x }\), the line with equation \(x = 1\), the positive \(x\)-axis and the line with equation \(x = a\) where \(a > 1\) A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is $$\pi \left( 1 - \frac { 1 } { a } \right)$$
2. Find the \(x\) coordinate of the centre of mass of \(S\).

8 Fig. 8 shows part of the graph of the function \(y = 5 x ( 2 x - 1 ) ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of S , the turning point of the curve.
2. Find the area of the shaded region enclosed between the curve and the \(x\)-axis.
3. Given that \(\mathrm { f } ( x ) = 5 x ( 2 x - 1 ) ^ { 3 }\), show that \(\mathrm { f } ( x + 0.5 ) = 40 x ^ { 3 } ( x + 0.5 )\).
4. Find \(\int _ { - \frac { 1 } { 2 } } ^ { 0 } 40 x ^ { 3 } ( x + 0.5 ) \mathrm { d } x\).
5. Explain, with the aid of a sketch, the connection between your answer to parts (ii) and (iv).

2. A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its acceleration is \(2 \sin \frac { 1 } { 2 } t \mathrm {~m} \mathrm {~s} ^ { - 2 }\), both measured in the direction of \(O x\). Given that \(v = 4\) when \(t = 0\),

1. find \(v\) in terms of \(t\),
2. calculate the distance travelled by \(P\) between the times \(t = 0\) and \(t = \frac { \pi } { 2 }\).

6. \includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-2_444_825_1571_516} The diagram shows the curve with equation \(y = \sqrt { \frac { x } { x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 3\).

1. Use Simpson's rule with six strips to estimate the area of the shaded region. The shaded region is rotated through four right angles about the \(x\)-axis.
2. Show that the volume of the solid formed is \(\pi ( 3 - \ln 4 )\).

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x\), giving your answer in exact form.

5 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

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

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

3 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\). [Hint: consider its reflection in \(y = x\).]

1 Show that \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }\).

2 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 \(x = 1\).

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).
[0pt] [3]

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

3 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

1 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).

1 Fig. 7 shows the curve \(y = _ { x - 1 }\). It has a minimum at the point P . The line \(l\) is an asymptote to the curve. \begin{figure}[h] \end{figure}

1. Write down the equation of the asymptote \(l\).
2. Find the coordinates of P .
3. Using the substitution \(u = x - 1\), show that the area of the region enclosed by the \(x\)-axis, the curve and the lines \(x = 2\) and \(x = 3\) is given by $$\int _ { 1 } ^ { 2 } \left( u + 2 + \frac { 4 } { u } \right) \mathrm { d } u$$ Evaluate this area exactly.
4. Another curve is defined by the equation \(\mathrm { e } ^ { y } = \frac { x ^ { 2 } + 3 } { x - 1 }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\) by differentiating implicitly. Hence find the gradient of this curve at the point where \(x = 2\).

2

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).