1.09f Trapezium rule: numerical integration

6

1. Use the trapezium rule with three intervals to find an approximation to \(\int _ { 1 } ^ { 4 } \frac { 6 } { 1 + \sqrt { x } } \mathrm {~d} x\). Give your answer correct to 5 significant figures.
2. Find the exact value of \(\int _ { 1 } ^ { 4 } 2 \mathrm { e } ^ { \frac { 1 } { 2 } x - 2 } \mathrm {~d} x\).
3. \includegraphics[max width=\textwidth, alt={}, center]{2d6fc4c5-70ec-4cd8-9b48-59d5ce0e39b7-11_556_805_262_705} The diagram shows the curves \(y = \frac { 6 } { 1 + \sqrt { x } }\) and \(y = 2 \mathrm { e } ^ { \frac { 1 } { 2 } x - 2 }\) which meet at a point with \(x\)-coordinate 4. The shaded region is bounded by the two curves and the line \(x = 1\). Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures.
4. State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region.

4

1. Use the trapezium rule with three intervals to show that the value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\) is approximately \(\ln 12\).
2. Use a graph of \(y = \ln x\) to show that \(\ln 12\) is an under-estimate of the true value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\).

4 \includegraphics[max width=\textwidth, alt={}, center]{9cf008d5-c15f-4491-9e4d-4bd070f896d5-06_446_832_260_653} The diagram shows part of the curve with equation \(y = \frac { 5 x } { 4 x ^ { 3 } + 1 }\). The shaded region is bounded by the curve and the lines \(x = 1 , x = 3\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of the maximum point.
2. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
3. State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.

4 \includegraphics[max width=\textwidth, alt={}, center]{c473f577-1e96-4d11-a0d5-cdfa4873c295-06_460_1445_260_349} The diagram shows the curve with equation \(y = \frac { x - 2 } { x ^ { 2 } + 8 }\). The shaded region is bounded by the curve and the lines \(x = 14\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence determine the exact \(x\)-coordinates of the stationary points.
2. Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{389df578-e7a7-4d19-9416-5e580d107717-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{1cd04df5-3fe3-4573-b880-d49262afd16a-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

6

1. 1. Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos 2 x \mathrm {~d} x = \frac { 1 } { 2 }\).
   2. By using an appropriate trigonometrical identity, find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 2 } x \mathrm {~d} x\).
2. 1. Use the trapezium rule with 2 intervals to estimate the value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec x d x\), giving your answer correct to 2 significant figures.
   2. Determine, by sketching the appropriate part of the graph of \(y = \sec x\), whether the trapezium rule gives an under-estimate or an over-estimate of the true value.

6 The equation of a curve is \(y = \frac { 1 } { 1 + \tan x }\).

1. Show, by differentiation, that the gradient of the curve is always negative.
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { 1 + \tan x } \mathrm {~d} x$$ giving your answer correct to 2 significant figures.
3. \includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-3_556_802_1384_708} The diagram shows a sketch of the curve for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\). State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).

5 \includegraphics[max width=\textwidth, alt={}, center]{34177829-f05d-449e-8881-5ab4d852c4ce-3_458_643_285_751} The diagram shows the part of the curve \(y = x \mathrm { e } ^ { - x }\) for \(0 \leqslant x \leqslant 2\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 2 } x \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).

6 \includegraphics[max width=\textwidth, alt={}, center]{08210e25-0f0e-405b-b72d-1bf989689b0a-3_641_865_264_641} The diagram shows the part of the curve \(y = \frac { \ln x } { x }\) for \(0 < x \leqslant 4\). The curve cuts the \(x\)-axis at \(A\) and its maximum point is \(M\).

1. Write down the coordinates of \(A\).
2. Show that the \(x\)-coordinate of \(M\) is e, and write down the \(y\)-coordinate of \(M\) in terms of e.
3. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 1 } ^ { 4 } \frac { \ln x } { x } \mathrm {~d} x$$ correct to 2 decimal places.
4. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii).

7 \includegraphics[max width=\textwidth, alt={}, center]{9d93ad8c-0a22-4de7-8342-387606e4e510-3_584_675_945_735} The diagram shows the part of the curve \(y = \mathrm { e } ^ { x } \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The curve meets the \(y\)-axis at the point \(A\). The point \(M\) is a maximum point.

1. Write down the coordinates of \(A\).
2. Find the \(x\)-coordinate of \(M\).
3. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } e ^ { x } \cos x d x$$ giving your answer correct to 2 decimal places.
4. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii).

3 \includegraphics[max width=\textwidth, alt={}, center]{b9556031-871d-4dd3-9523-e3438a41339f-2_451_775_559_683} The diagram shows the curve \(y = \frac { 1 } { 1 + \sqrt { } x }\) for values of \(x\) from 0 to 2 .

1. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 2 } \frac { 1 } { 1 + \sqrt { } x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (i).

2 \includegraphics[max width=\textwidth, alt={}, center]{f55e1431-9443-40d7-bec0-6f5c1d9fa2d7-2_531_949_431_598} The diagram shows part of the curve \(y = x \mathrm { e } ^ { - x }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 2 , x = 3\) and \(y = 0\).

1. Use the trapezium rule with two intervals to estimate the area of \(R\), giving your answer correct to 2 decimal places.
2. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of \(R\).

2 \includegraphics[max width=\textwidth, alt={}, center]{af7c2b6e-1293-4744-8ea0-927fea5ab4ec-2_531_949_431_598} The diagram shows part of the curve \(y = x \mathrm { e } ^ { - x }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 2 , x = 3\) and \(y = 0\).

1. Use the trapezium rule with two intervals to estimate the area of \(R\), giving your answer correct to 2 decimal places.
2. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of \(R\).

2 \includegraphics[max width=\textwidth, alt={}, center]{2c27f384-5289-4c1b-9199-6b4c6ac81e38-2_645_750_429_699} The diagram shows the curve \(y = \sqrt { } \left( 1 + x ^ { 3 } \right)\). Region \(A\) is bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\). Region \(B\) is bounded by the curve and the lines \(x = 0\) and \(y = 3\).

1. Use the trapezium rule with two intervals to find an approximation to the area of region \(A\). Give your answer correct to 2 decimal places.
2. Deduce an approximation to the area of region \(B\) and explain why this approximation underestimates the true area of region \(B\).

2 \includegraphics[max width=\textwidth, alt={}, center]{ee420db2-bef4-4c2b-8dd2-c8f439dd561e-2_645_750_429_699} The diagram shows the curve \(y = \sqrt { } \left( 1 + x ^ { 3 } \right)\). Region \(A\) is bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\). Region \(B\) is bounded by the curve and the lines \(x = 0\) and \(y = 3\).

1. Use the trapezium rule with two intervals to find an approximation to the area of region \(A\). Give your answer correct to 2 decimal places.
2. Deduce an approximation to the area of region \(B\) and explain why this approximation underestimates the true area of region \(B\).

6

1. Show that \(\int _ { 6 } ^ { 16 } \frac { 6 } { 2 x - 7 } \mathrm {~d} x = \ln 125\).
2. Use the trapezium rule with four intervals to find an approximation to $$\int _ { 1 } ^ { 17 } \log _ { 10 } x d x$$ giving your answer correct to 3 significant figures.

3

1. Find \(\int 4 \cos \left( \frac { 1 } { 3 } x + 2 \right) \mathrm { d } x\).
2. Use the trapezium rule with three intervals to find an approximation to $$\int _ { 0 } ^ { 12 } \sqrt { } \left( 4 + x ^ { 2 } \right) \mathrm { d } x$$ giving your answer correct to 3 significant figures.

6

1. Find \(\int \frac { 4 + \mathrm { e } ^ { x } } { 2 \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).
2. Without using a calculator, find \(\int _ { 2 } ^ { 10 } \frac { 1 } { 2 x + 5 } \mathrm {~d} x\), giving your answer in the form \(\ln k\).
3. \includegraphics[max width=\textwidth, alt={}, center]{a07e6d2f-ded1-4c62-957b-41fb94b46a2d-3_446_755_580_735} The diagram shows the curve \(y = \log _ { 10 } ( x + 2 )\) for \(0 \leqslant x \leqslant 6\). The region bounded by the curve and the lines \(x = 0 , x = 6\) and \(y = 0\) is denoted by \(R\). Use the trapezium rule with 2 strips to find an estimate of the area of \(R\), giving your answer correct to 1 decimal place.

6

1. Find \(\int \frac { 4 + \mathrm { e } ^ { x } } { 2 \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).
2. Without using a calculator, find \(\int _ { 2 } ^ { 10 } \frac { 1 } { 2 x + 5 } \mathrm {~d} x\), giving your answer in the form \(\ln k\).
3. \includegraphics[max width=\textwidth, alt={}, center]{f85c4010-17b1-441c-ae8a-e77573d1b0c3-3_446_755_580_735} The diagram shows the curve \(y = \log _ { 10 } ( x + 2 )\) for \(0 \leqslant x \leqslant 6\). The region bounded by the curve and the lines \(x = 0 , x = 6\) and \(y = 0\) is denoted by \(R\). Use the trapezium rule with 2 strips to find an estimate of the area of \(R\), giving your answer correct to 1 decimal place.

6 \includegraphics[max width=\textwidth, alt={}, center]{6295873e-7db4-4e7e-8dcd-912ad9c41675-06_561_542_260_799} The diagram shows the curve \(y = \tan 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\). The shaded region is bounded by the curve and the lines \(x = \frac { 1 } { 6 } \pi\) and \(y = 0\).

1. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region, giving your answer correct to 3 significant figures.
2. Find the exact volume of the solid formed when the shaded region is rotated completely about the \(x\)-axis.

2 \includegraphics[max width=\textwidth, alt={}, center]{208eab3e-a78c-43b4-918f-a9efc9b4f47a-2_508_586_450_776} The diagram shows a sketch of the curve \(y = \frac { 1 } { 1 + x ^ { 3 } }\) for values of \(x\) from - 0.6 to 0.6 .

1. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { - 0.6 } ^ { 0.6 } \frac { 1 } { 1 + x ^ { 3 } } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Explain, with reference to the diagram, why the trapezium rule may be expected to give a good approximation to the true value of the integral in this case.

2 \includegraphics[max width=\textwidth, alt={}, center]{0f73e750-18a0-49ad-b4cb-fd6d14f0789e-2_531_700_395_719} The diagram shows the curve \(y = \sqrt { } \left( 1 + 2 \tan ^ { 2 } x \right)\) for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\).

1. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sqrt { } \left( 1 + 2 \tan ^ { 2 } x \right) \mathrm { d } x$$ giving your answer correct to 2 decimal places.
2. The estimate found in part (i) is denoted by \(E\). Explain, without further calculation, whether another estimate found using the trapezium rule with six intervals would be greater than \(E\) or less than \(E\).

2 Use the trapezium rule with three intervals to find an approximation to $$\int _ { 0 } ^ { 3 } \left| 3 ^ { x } - 10 \right| \mathrm { d } x$$

1 Use the trapezium rule with three intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \ln ( 1 + \sin x ) \mathrm { d } x$$ giving your answer correct to 2 decimal places.