1.08e Area between curve and x-axis: using definite integrals

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 the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

9 The curve in Fig. 9.1 has equation \(\sqrt { x } + \sqrt { y } = 1\). \begin{figure}[h] \end{figure}

1. Show that this is part, but not all of the curve \(y = 1 - 2 \sqrt { x } + x\). Sketch the full curve \(y = 1 - 2 \sqrt { x } + x\).
2. Fig.9.2 shows a star shape made up of four parts, one of which is given in part (i) above. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_380_681_1197_651} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} For each of the sections of the shape labelled \(\mathrm { A } , \mathrm { B }\) and C , state the equation of the curve and the domain.
3. The shape shown in Fig.9.2 is made into that in Fig. 10.3 by stretching the part of the figure for which \(y > 0\) by a scale factor of 2 . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_405_686_1996_605} \captionsetup{labelformat=empty} \caption{Fig. 9.3}
   \end{figure} Find the area of this shape.

2 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

4 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
4. Find the exact area of the shaded region between the line OP and the curve.

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 \(x = 2\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x _ { 2 } } }\). \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. (A) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

1 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 \(x = 4.5\).

2 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

3 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 \(x = 1\).

2 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\). The endpoints of the curve are \(\mathrm { P } ( - \pi , 1 )\) and \(\mathrm { Q } ( \pi , 3 )\), and \(\mathrm { f } ( x ) = a + \sin b x\), where \(a\) and \(b\) are constants. \begin{figure}[h] \end{figure}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac { 1 } { 2 }\).
2. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,2 ). Show that there is no point on the curve at which the gradient is greater than this.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range. Write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 2,0 )\).
4. Find the area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h] \end{figure}

1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).

1 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

2 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
4. Find the exact area of the shaded region between the line OP and the curve. END OF QUESTION PAPER

1 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P . \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

3 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).

2 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

7 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.

2 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B, as shown. The curves and the line \(y = x\) meet at the point C . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
2. Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
3. Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e .
4. Use integration by parts to find \(\int \ln x \mathrm {~d} x\). Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
5. Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .