Over/underestimate justification with graph

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

1. Use the trapezium rule with 3 intervals to estimate the value of $$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 2 } { 3 } \pi } \operatorname { cosec } x d x$$ giving your answer correct to 2 decimal places.
2. Using a sketch of the graph of \(y = \operatorname { cosec } x\), explain whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).

8 \includegraphics[max width=\textwidth, alt={}, center]{5b5ed7d1-028e-4f9a-ae9e-26071d0df678-14_604_497_262_822} The diagram shows the graph of \(y = \sec x\) for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Use the trapezium rule with 2 intervals to estimate the value of \(\int _ { 0 } ^ { 1.2 } \sec x \mathrm {~d} x\), giving your answer correct to 2 decimal places.
2. Explain, with reference to the diagram, whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).
3. \(P\) is the point on the part of the curve \(y = \sec x\) for \(0 \leqslant x < \frac { 1 } { 2 } \pi\) at which the gradient is 2 . By first differentiating \(\frac { 1 } { \cos x }\), find the \(x\)-coordinate of \(P\), giving your answer correct to 3 decimal places.

3

1. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Using a sketch of the graph of \(y = \sec x\) for \(0 \leqslant x \leqslant \frac { 1 } { 3 } \pi\), explain whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (i).

3 \includegraphics[max width=\textwidth, alt={}, center]{322eb555-d40a-460c-8c71-5780f5772bcd-2_535_1041_573_552} The diagram shows the curve \(y = x - 2 \ln x\) and its minimum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 2 } ^ { 5 } ( x - 2 \ln x ) \mathrm { d } x$$ 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

1. Find \(\int ( \sin x - \cos x ) ^ { 2 } \mathrm {~d} x\).
2. 1. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \operatorname { cosec } x d x$$ giving your answer correct to 3 decimal places.
   2. Using a sketch of the graph of \(y = \operatorname { cosec } x\) for \(0 < x \leqslant \frac { 1 } { 2 } \pi\), explain whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (i).

4 \includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-2_543_857_1155_644} The diagram shows the curve \(\mathrm { y } = \sqrt { 4 \mathrm { X } + 1 }\).

1. Use the trapezium rule, with strips of width 0.5 , to find an approximate value for the area of the region bounded by the curve \(y = \sqrt { 4 x + 1 }\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 3\). Give your answer correct to 3 significant figures.
2. State with a reason whether this approximation is an under-estimate or an over-estimate.

4 \begin{figure}[h] \end{figure} Fig. 4 shows a curve which passes through the points shown in the following table. Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve, the lines \(x = 1\) and \(x = 4\), and the \(x\)-axis. State, with a reason, whether the trapezium rule gives an overestimate or an underestimate of the area of this region.

2 \begin{figure}[h] \end{figure} Fig. 4 shows a curve which passes through the points shown in the following table. Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve, the lines \(x = 1\) and \(x = 4\), and the \(x\)-axis. State, with a reason, whether the trapezium rule gives an overestimate or an underestimate of the area of this region.
[0pt] [5]

3 \includegraphics[max width=\textwidth, alt={}, center]{c52fe7e9-0442-4b3e-b924-2e5e4b3e98f5-02_510_791_991_678} The diagram shows the curve \(y = \sqrt { x - 3 }\).

1. Use the trapezium rule, with 4 strips each of width 0.5 , to find an approximate value for the area of the region bounded by the curve, the \(x\)-axis and the line \(x = 5\). Give your answer correct to 3 significant figures.
2. State, with a reason, whether this approximation is an underestimate or an overestimate.

2 \includegraphics[max width=\textwidth, alt={}, center]{ad3083ae-caa6-42d8-a1f2-e984150cb104-2_536_917_1016_577} The diagram shows the curve \(y = \log _ { 10 } ( 2 x + 1 )\).

1. Use the trapezium rule with 4 strips each of width 1.5 to find an approximation to the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 4\) and \(x = 10\). Give your answer correct to 3 significant figures.
2. Explain why this approximation is an under-estimate.

4 Fig. 4 shows the graph of \(y = \sqrt { 1 + x ^ { 3 } }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with \(h = 0.5\) to find an estimate of \(\int _ { - 1 } ^ { 0 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your answer correct to 6 decimal places.
2. State whether your answer to part (a) is an under-estimate or an over-estimate, justifying your answer.

4 A spreadsheet is used to approximate \(\int _ { a } ^ { b } f ( x ) d x\) using the midpoint rule with 1 strip. The output is shown in the table below. The formula in cell C4 is \(= \mathrm { B } 4 \wedge ( 1 / \mathrm { B } 4 )\).
The formula in cell D4 is \(= 0.5 ^ { * } \mathrm { C } 4\).

1. Write the integral in standard mathematical notation. A graph of \(y = f ( x )\) is included in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-04_789_1004_1199_235}
2. Explain whether 0.65518535 is an over-estimate or an under-estimate of \(\int _ { a } ^ { b } f ( x ) d x\).

6 The diagram below shows part of the graph of \(y = \mathrm { f } ( x )\) The line \(T P Q\) is a tangent to the graph of \(y = \mathrm { f } ( x )\) at the point \(P \left( \frac { a + b } { 2 } , \mathrm { f } \left( \frac { a + b } { 2 } \right) \right)\) The points \(S ( a , 0 )\) and \(T\) lie on the line \(x = a\) The points \(Q\) and \(R ( b , 0 )\) lie on the line \(x = b\) \includegraphics[max width=\textwidth, alt={}, center]{74b8239a-1f46-45e7-ad20-2dce7bf4baf6-05_748_696_669_671} Sharon uses the mid-ordinate rule with one strip to estimate the value of the integral \(\int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x\) By considering the area of the trapezium QRST, state, giving reasons, whether you would expect Sharon's estimate to be an under-estimate or an over-estimate.

\includegraphics{figure_3} Figure 3 shows a sketch of the curve with equation \(y = 2x^2 - x\). The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the line with equation \(x = -0.5\), the \(x\)-axis and the line with equation \(x = 1.5\).

1. The trapezium rule with four strips is used to find an estimate for the area of \(R\). Explain whether the estimate for R is an underestimate or overestimate to the true value for the area of \(R\). [1]

The estimate for R is found to be 2.58. Using this value, and showing your working,

1. estimate the value of \(\int_{-0.5}^{1.5} (2x^2 + 1 + 2x) \, dx\). [3]