Definite integrals and areas - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 19 exam-style questions on Definite integrals and areas for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
2–11 marks each
Instant marking
No sign-up needed

19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Find the exact value of
\[\lim_{\delta x \to 0} \sum_{x = \frac{\pi}{6}}^{\frac{\pi}{3}} 2\sin 3x\,\delta x\]

Model answer

\(\displaystyle\lim_{\delta x \to 0} \sum_{x = \frac{\pi}{6}}^{\frac{\pi}{3}} 2\sin 3x\,\delta x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 2\sin 3x\,dx\)
\(= \left[-\dfrac{2}{3}\cos 3x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)
\(= -\dfrac{2}{3}\cos\pi - \left(-\dfrac{2}{3}\cos\dfrac{\pi}{2}\right) = \dfrac{2}{3} - 0 = \dfrac{2}{3}\)

Explanation

As \(\delta x \to 0\), the sum of \(\mathrm{f}(x)\,\delta x\) becomes the definite integral of \(\mathrm{f}(x)\) over the same interval, so the first mark is for writing that integral with its limits and \(dx\). Integrating \(\sin 3x\) gives \(-\dfrac{1}{3}\cos 3x\), and the limits are in radians. Common errors are leaving out the limits or the \(dx\), losing the minus sign, and working in degrees.

Mark scheme

1 mark
B1 Writes the limit as \(\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 2\sin 3x\,dx\), including the limits and the \(dx\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands working. Condone \(\delta x\) in place of \(dx\). The integral may be implied by correct integration followed by substitution of the correct limits.
1 mark
M1 Integrates \(2\sin 3x\) to \(k\cos 3x\), \(k\) ≠ 0, and substitutes both limits
The correct integral is \(-\dfrac{2}{3}\cos 3x\). Any non-zero \(k\) earns the method mark.
1 mark
A1 \(\dfrac{2}{3}\) cso
Needs \(\cos\pi = -1\) and \(\cos\dfrac{\pi}{2} = 0\), seen or implied: \(-\dfrac{2}{3}(-1) - \left(-\dfrac{2}{3} \times 0\right) = \dfrac{2}{3}\). Accept \(0.\dot{6}\) if the recurring dot is clearly shown, but not 0.667. Working in degrees scores A0.

Question 2

6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The curve \(C\) has equation \(y = x^3 - 5x^2 + 4x + 9\) and the line \(l\) has equation \(y = 2x + 1\).

Figure 1 shows \(C\) and \(l\), which meet at the points \(A\), \(B\) and \(D\). The point \(A\) has \(x\)-coordinate \(-1\). The finite region \(R_1\), shown shaded in Figure 1, is bounded by \(C\) and \(l\) between \(A\) and \(B\). The finite region \(R_2\), also shown shaded, is bounded by \(C\) and \(l\) between \(B\) and \(D\).

Using algebra, find the \(x\)-coordinates of \(B\) and \(D\).

Use algebraic integration to find the exact area of \(R_2\).

Figure 1: the curve C and the straight line l, which cross at three points A, B and D, with B between A and D. The region R1 between the curve and the line from A to B, where the curve is above the line, is shaded, and so is the region R2 between them from B to D, where the line is above the curve.

Model answer

\(x^3 - 5x^2 + 4x + 9 = 2x + 1 \Rightarrow x^3 - 5x^2 + 2x + 8 = 0\)
\(x = -1\) is a root, so \((x + 1)\) is a factor: \(x^3 - 5x^2 + 2x + 8 = (x + 1)\left(x^2 - 6x + 8\right) = (x + 1)(x - 2)(x - 4)\)
So \(B\) has \(x = 2\) and \(D\) has \(x = 4\).
Between \(B\) and \(D\), \(l\) lies above \(C\):
Area of \(R_2 = \displaystyle\int_2^4 \left((2x + 1) - (x^3 - 5x^2 + 4x + 9)\right)dx = \int_2^4 \left(-x^3 + 5x^2 - 2x - 8\right)dx\)
\(= \left[-\dfrac{x^4}{4} + \dfrac{5x^3}{3} - x^2 - 8x\right]_2^4\)
\(= \left(-64 + \dfrac{320}{3} - 16 - 32\right) - \left(-4 + \dfrac{40}{3} - 4 - 16\right)\)
\(= -\dfrac{16}{3} - \left(-\dfrac{32}{3}\right) = \dfrac{16}{3}\)

Explanation

The \(x\)-coordinates come from \(x^3 - 5x^2 + 2x + 8 = 0\); since \(x = -1\) is a root, \((x + 1)\) is a factor, and the quadratic factor \(x^2 - 6x + 8\) gives \(x = 2\) and \(x = 4\). On \(2 \leqslant x \leqslant 4\) the line is above the curve, so integrate (line minus curve) between these limits. Common errors are using \(-1\) and 4 as the limits, which mixes \(R_1\) with \(R_2\), sign slips when subtracting the curve from the line, and leaving the area negative.

Mark scheme

1 mark
M1 Sets \(x^3 - 5x^2 + 4x + 9 = 2x + 1\) to obtain \(x^3 - 5x^2 + 2x + 8 = 0\), uses \((x + 1)\) as a factor (by division or comparing coefficients) to obtain the quadratic factor \(x^2 - 6x + 8\), and solves it by factorising, the formula or completing the square. An answer with no working scores 0 marks.
An answer of \(x = 2\), \(x = 4\) and area \(\dfrac{16}{3}\) with no working scores 0 marks: the question demands working. Roots taken straight from a calculator's equation solver, without a quadratic factor being seen, score M0.
1 mark
A1 \(x = 2\) (at \(B\)) and \(x = 4\) (at \(D\))
Both values are needed. Ignore \(x = -1\).
1 mark
M1 Integrates \(\pm\left((2x + 1) - (x^3 - 5x^2 + 4x + 9)\right)\), or each function separately: at least one power increased by 1
Not dependent on the first M mark. Integrating each function separately and subtracting the results later is fine.
1 mark
A1 \(\pm\left(\dfrac{x^4}{4} - \dfrac{5x^3}{3} + x^2 + 8x\right)\) oe
1 mark
M1 (dM1) Substitutes the limits 4 and 2 into their integrated expression and subtracts
Depends on the second M mark. The limits must be their \(x\)-coordinates of \(B\) and \(D\); using \(-1\) and 4 (which combines \(R_1\) and \(R_2\)) is M0.
1 mark
A1 Area of \(R_2 = \dfrac{16}{3}\) cso
If \(\displaystyle\int_2^4 (x^3 - 5x^2 + 2x + 8)\,dx = -\dfrac{16}{3}\) is found, the final answer must be the positive value. Accept \(5\tfrac{1}{3}\); 5.33 on its own scores A0.

Question 3

7 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The curve \(C\) has equation
\[y = 3 - \mathrm{e}^{\frac{x}{2}}, \quad 0 \leqslant x \leqslant 4\]
and cuts the \(x\)-axis at \(A\). Two regions are shaded in Figure 1: \(R_1\), enclosed by \(C\), the \(y\)-axis and the \(x\)-axis, which lies above the \(x\)-axis; and \(R_2\), enclosed by \(C\), the \(x\)-axis and the vertical line \(x = 4\), which lies below the \(x\)-axis.

Find the exact value of the sum of the areas of \(R_1\) and \(R_2\), giving your answer in the form \(a\ln 3 + b\mathrm{e}^2 + c\), where \(a\), \(b\) and \(c\) are integers.

Figure 1: the curve C starts on the positive y-axis and falls more and more steeply, crossing the x-axis at A and continuing below the axis to the vertical line x = 4. The region R1 between the curve and the axes, above the x-axis, is shaded, and so is the region R2 between the curve, the x-axis and the line x = 4, below the x-axis.

Model answer

\(3 - \mathrm{e}^{\frac{x}{2}} = 0 \Rightarrow \mathrm{e}^{\frac{x}{2}} = 3 \Rightarrow x = 2\ln 3\) at \(A\)
\(\displaystyle\int \left(3 - \mathrm{e}^{\frac{x}{2}}\right)dx = 3x - 2\mathrm{e}^{\frac{x}{2}}\)
\(R_1\): \(\left[3x - 2\mathrm{e}^{\frac{x}{2}}\right]_0^{2\ln 3} = (6\ln 3 - 6) - (0 - 2) = 6\ln 3 - 4\)
\(R_2\): \(\left[3x - 2\mathrm{e}^{\frac{x}{2}}\right]_{2\ln 3}^{4} = \left(12 - 2\mathrm{e}^2\right) - (6\ln 3 - 6) = 18 - 2\mathrm{e}^2 - 6\ln 3\)
This is negative because \(R_2\) is below the \(x\)-axis, so the area of \(R_2\) is \(2\mathrm{e}^2 + 6\ln 3 - 18\).
Total area \(= (6\ln 3 - 4) + \left(2\mathrm{e}^2 + 6\ln 3 - 18\right) = 12\ln 3 + 2\mathrm{e}^2 - 22\)

Explanation

The curve crosses the \(x\)-axis where \(\mathrm{e}^{\frac{x}{2}} = 3\), at \(x = 2\ln 3\), so the two regions must be integrated separately. The integral over \(R_2\) is negative because that region is below the axis, so its magnitude is added to the area of \(R_1\). A common error is to integrate from 0 to 4 in one step, which subtracts the area of \(R_2\) instead of adding it; another is integrating \(\mathrm{e}^{\frac{x}{2}}\) as \(\dfrac{1}{2}\mathrm{e}^{\frac{x}{2}}\).

Mark scheme

1 mark
B1 The \(x\)-coordinate of \(A\) is \(2\ln 3\) (or \(\ln 9\)). An answer with no working scores 0 marks.
An answer of \(12\ln 3 + 2\mathrm{e}^2 - 22\) with no working scores 0 marks: the question demands working. The value may be seen as a limit.
1 mark
M1 Integrates \(3 - \mathrm{e}^{\frac{x}{2}}\): \(\mathrm{e}^{\frac{x}{2}} \to k\mathrm{e}^{\frac{x}{2}}\), \(k\) ≠ 0
For example \(\dfrac{1}{2}\mathrm{e}^{\frac{x}{2}}\) earns this mark but not the next.
1 mark
A1 \(3x - 2\mathrm{e}^{\frac{x}{2}}\)
1 mark
M1 (dM1) Substitutes the limits 0 and their \(2\ln 3\) for \(R_1\), and their \(2\ln 3\) and 4 for \(R_2\), into their integrated expression and subtracts in each case
Depends on the M mark for integrating. For \(R_1\): \((6\ln 3 - 6) - (0 - 2)\). For \(R_2\): \((12 - 2\mathrm{e}^2) - (6\ln 3 - 6) = 18 - 2\mathrm{e}^2 - 6\ln 3\), which is negative.
1 mark
A1 Area of \(R_1 = 6\ln 3 - 4\)
1 mark
M1 (ddM1) Adds the area of \(R_1\) to the magnitude of the integral for \(R_2\), i.e. area of \(R_2 = 2\mathrm{e}^2 + 6\ln 3 - 18\)
Depends on both previous M marks. Integrating from 0 to 4 in one go, giving \(14 - 2\mathrm{e}^2\) (the two regions partly cancel), scores M0 here.
1 mark
A1 \(12\ln 3 + 2\mathrm{e}^2 - 22\) cso
\(a = 12\), \(b = 2\), \(c = -22\). An answer containing \(\ln 9\) must be written in the required form, \(12\ln 3\), for this mark. A decimal (5.96) on its own scores A0.

Question 4

9 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Figure 1 shows the side profile of a slope at a ski centre. The profile is modelled by the curve \(C\) with equation
\[y = 60\mathrm{e}^{-\frac{x}{50}}, \quad 0 \leqslant x \leqslant 100\]
where \(x\) metres is the horizontal distance from the top of the slope and \(y\) metres is the height above the base of the slope. The point \(A\) on \(C\), where \(x = 0\), is the top of the slope and the point \(B\) on \(C\), where \(x = 100\), is the bottom.

To make a straight run for beginners, the hollow between \(C\) and the straight line \(AB\) is to be filled with snow. The finite region \(R\), shown shaded in Figure 1, is bounded by \(C\) and the line \(AB\).

Show that the area of \(R\) is \(6000\mathrm{e}^{-2}\) m\(^2\).

The run is 40 m wide and the cross-section is the same across its width. Find, according to the model, the volume of snow needed. Give your answer in m\(^3\) to 3 significant figures.

Give one reason why the volume of snow needed in practice may be different from your answer.

Model answer

At \(B\), \(y = 60\mathrm{e}^{-2}\).
Area under \(AB\) (a trapezium) \(= \dfrac{1}{2}\left(60 + 60\mathrm{e}^{-2}\right) \times 100 = 3000 + 3000\mathrm{e}^{-2}\)
Area under \(C = \displaystyle\int_0^{100} 60\mathrm{e}^{-\frac{x}{50}}\,dx = \left[-3000\mathrm{e}^{-\frac{x}{50}}\right]_0^{100} = -3000\mathrm{e}^{-2} + 3000\)
Area of \(R = \left(3000 + 3000\mathrm{e}^{-2}\right) - \left(3000 - 3000\mathrm{e}^{-2}\right) = 6000\mathrm{e}^{-2}\) m\(^2\)
Volume \(= 40 \times 6000\mathrm{e}^{-2} = 240000\mathrm{e}^{-2} = 32500\) m\(^3\) (3 s.f.)
In practice snow settles and compacts, so more snow than this would be needed.

Explanation

The curve is below the chord \(AB\), so the area of \(R\) is the area under \(AB\), a trapezium with parallel sides 60 and \(60\mathrm{e}^{-2}\), minus the area under the curve. Integrating \(60\mathrm{e}^{-\frac{x}{50}}\) gives \(-3000\mathrm{e}^{-\frac{x}{50}}\), and the volume is the area times the width. Common errors are using a triangle instead of a trapezium and integrating the exponential with a factor of \(-\dfrac{1}{50}\) instead of \(-50\).

Mark scheme

1 mark
B1 The height of \(B\) is \(60\mathrm{e}^{-2}\) and the area under \(AB\), a trapezium, is \(\dfrac{1}{2}\left(60 + 60\mathrm{e}^{-2}\right) \times 100 = 3000 + 3000\mathrm{e}^{-2}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the area is a given answer and working is demanded.
1 mark
M1 Integrates \(60\mathrm{e}^{-\frac{x}{50}}\) to \(k\mathrm{e}^{-\frac{x}{50}}\), \(k\) ≠ 0
1 mark
A1 \(-3000\mathrm{e}^{-\frac{x}{50}}\)
1 mark
M1 (dM1) Substitutes the limits 100 and 0 and subtracts: \(\displaystyle\int_0^{100} 60\mathrm{e}^{-\frac{x}{50}}\,dx = 3000 - 3000\mathrm{e}^{-2}\)
Depends on the previous M mark.
1 mark
M1 Area of \(R\) = area under \(AB\) minus area under \(C\)
Alternative: integrate (line minus curve), with the line \(y = 60 - 0.6\left(1 - \mathrm{e}^{-2}\right)x\); the marks are awarded in the same order.
1 mark
A1* \(\left(3000 + 3000\mathrm{e}^{-2}\right) - \left(3000 - 3000\mathrm{e}^{-2}\right) = 6000\mathrm{e}^{-2}\) cso
A given answer: both areas must be seen in exact form.
1 mark
M1 Volume \(= 40 \times 6000\mathrm{e}^{-2}\)
1 mark
A1 awrt 32500 m\(^3\), i.e. [32450, 32550)
\(240000\mathrm{e}^{-2} = 32480.5\ldots\); accept 32480 or more accurate values. Units are not required.
1 mark
B1 A valid reason, e.g. snow settles or compacts (or melts), so more is needed; the real ground does not follow the curve exactly; the cross-section is not the same across the whole width
Do not accept vague answers such as "the model is inaccurate" or "rounding errors".

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.