1.08i Integration by parts

8 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

2 Find \(\int x \mathrm { e } ^ { 3 x } \mathrm {~d} x\).

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.

5 Find \(\int _ { 2 } ^ { 3 } x \mathrm { e } ^ { 2 x } \mathrm {~d} x\), giving your answer to 1 decimal place.

6 Find \(\frac { \mathrm { d } } { \mathrm { d } x } ( x \ln x )\) and hence or otherwise find the value of \(\int _ { 2 } ^ { 3 } \ln x \mathrm {~d} x\), giving your answer in the form \(\ln a + b\), where \(a\) and \(b\) are to be determined.

8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the points where the curve cuts the axes.
2. Find the coordinates of the stationary point, S , on the curve.
3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
4. Sketch the curve in the domain \(- 3 < x < 3\).
5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

6

1. Find \(\int x \cos 2 x d x\).
2. Using the substitution \(u = x ^ { 2 } + 1\), or otherwise, find the exact value of \(\int _ { 2 } ^ { 3 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x\).

1 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h] \end{figure}

1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
2. Find the exact coordinates of the turning point P .
3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\) - and \(y\)-axes.

4

1. Differentiate \(\frac { \ln x } { x ^ { 2 } }\), simplifying your answer.
2. Using integration by parts, show that \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c\).

3

1. Differentiate \(x \cos 2 x\) with respect to \(x\).
2. Integrate \(x \cos 2 x\) with respect to \(x\).

3

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]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_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 \(x = 1\).

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.

3 A curve is defined by the equation \(y = 2 x \ln ( 1 + x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

1 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

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 \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

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 .

6 Fig. 8 shows part of the curve \(y = x \cos 3 x\). The curve crosses the \(x\)-axis at \(\mathrm { O } , \mathrm { P }\) and Q . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P and Q .
2. Find the exact gradient of the curve at the point P . Show also that the turning points of the curve occur when \(x \tan 3 x = \frac { 1 } { 3 }\).
3. Find the area of the region enclosed by the curve and the \(x\)-axis between O and P , giving your answer in exact form.

1 Find the exact value of \(\int ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x\).

4 Show that \(\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos \frac { 1 } { 2 } x \mathrm {~d} x = \frac { \sqrt { 2 } } { 2 } \pi + 2 \sqrt { 2 } - 4\).
[0pt] [5]

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 \(x = 1\).

3

1. Differentiate \(\frac { \ln x } { x ^ { 2 } }\), simplifying your answer.
2. Using integration by parts, show that \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c\).

2 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.

1 Evaluate \(\int _ { 1 } ^ { 2 } x ^ { 2 } \ln x \mathrm {~d} x\), giving your answer in an exact form.

3 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

1 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 .