Edexcel A level Maths (9MA0) · Pure mathematics › Integration
Practise Areas between curves and parametric areas. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy7 marks
The diagram shows part of the curve \(C\) with equation \(y = 6x - x^{2}\) and the straight line \(l\) with equation \(y = x\). The line \(l\) meets \(C\) at the origin and at the point \(A\). The region \(R\), shown shaded, is bounded by \(C\) and \(l\). In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the coordinates of \(A\).[3]
(b) Use integration to find the exact area of \(R\).[4]
Show the answer and mark scheme
(a)Answer: \(A(5, 5)\)
M1 for equating \(6x - x^{2} = x\)
A1 for \(x = 5\)
A1 for \(A(5, 5)\)
Worked solution: \(6x - x^{2} = x \Rightarrow x^{2} - 5x = 0 \Rightarrow x(x - 5) = 0\), so at \(A\), \(x = 5\) and \(y = 5\).
(b)Answer: \(\frac{125}{6}\)
M1 for \(\int_{0}^{5} \left((6x - x^{2}) - x\right) \,\mathrm{d}x\) (curve minus line, or area under curve minus area of triangle)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for \(\left[\frac{5}{2}x^{2} - \frac{1}{3}x^{3}\right]_{0}^{5}\)
A1 for \(\frac{125}{6}\)
Worked solution: Area \(= \int_{0}^{5} (5x - x^{2}) \,\mathrm{d}x = \left[\frac{5}{2}x^{2} - \frac{1}{3}x^{3}\right]_{0}^{5}\) \(= \left(\frac{125}{6}\right) - \left(0\right) = \frac{125}{6}\). (Equivalently, area under \(C\) minus the area of the triangle below \(OA\).)
Question 2Medium9 marks
The diagram shows part of the curve \(C\) with equation \(y = -x^{2} - x - 1\) and the line \(l\) with equation \(y = -3\). The line meets \(C\) at the points \(A\) and \(B\). The region \(R\), shown shaded, is bounded by \(C\) and \(l\). In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the coordinates of \(A\) and \(B\).[4]
(b) Find the exact area of \(R\).[5]
Show the answer and mark scheme
(a)Answer: \(A(-2, -3)\), \(B(1, -3)\)
M1 for equating the equations of \(C\) and \(l\)
A1 for a correct 3TQ, e.g. \(x^{2} + x - 2 = 0\)
M1 for solving their 3TQ
A1 for \(A(-2, -3)\) and \(B(1, -3)\)
Worked solution: \(-x^{2} - x - 1 = -3 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow (x + 2)(x - 1) = 0\). So \(x = -2\) or \(x = 1\), giving \(A(-2, -3)\) and \(B(1, -3)\).
(b)Answer: \(\frac{9}{2}\)
M1 for an integral of the form \(\int (\text{upper} - \text{lower}) \,\mathrm{d}x\) with their limits
A1 for the integrand \(-x^{2} - x + 2\)
M1 for \(x^{n} \to x^{n + 1}\)
M1 for substituting their limits and subtracting
A1 for \(\frac{9}{2}\)
Worked solution: The curve is above the line for \(-2 \lt x \lt 1\), so area \(= \int_{-2}^{1} (-x^{2} - x + 2) \,\mathrm{d}x = \left[-\frac{1}{3}x^{3} - \frac{1}{2}x^{2} + 2x\right]_{-2}^{1}\) \(= \left(\frac{7}{6}\right) - \left(-\frac{10}{3}\right) = \frac{9}{2}\).
Question 3Hard9 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Show that, for \(0 \le x \le \pi\), the curves \(y = \sin 2x\) and \(y = \sin x\) meet where \(x = 0\), \(x = \frac{\pi}{3}\) and \(x = \pi\).[2]
(b) Find the area of the region enclosed between the two curves for \(0 \le x \le \frac{\pi}{3}\).[3]
(c) A student claims that the total area enclosed between the two curves for \(0 \le x \le \pi\) is found as follows: \(\int_{0}^{\pi} (\sin 2x - \sin x)\,\mathrm{d}x = \left[-\frac{1}{2}\cos 2x + \cos x\right]_{0}^{\pi} = \left(-\frac{1}{2} - 1\right) - \left(-\frac{1}{2} + 1\right) = -2\), so the area is 2. Explain the error in the student's method and find the correct total area.[4]
Show the answer and mark scheme
(a)Answer: \(2\sin x\cos x = \sin x \Rightarrow \sin x(2\cos x - 1) = 0\), so \(\sin x = 0\) (\(x = 0, \pi\)) or \(\cos x = \frac{1}{2}\) (\(x = \frac{\pi}{3}\)).
M1 for \(\sin x(2\cos x - 1) = 0\)
A1* for all three values and no others in the interval
Worked solution: \(\sin 2x = \sin x \Rightarrow 2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1) = 0\). \(\sin x = 0\): \(x = 0\) or \(\pi\). \(\cos x = \frac{1}{2}\): \(x = \frac{\pi}{3}\).
(b)Answer: \(\frac{1}{4}\)
M1 for \(\int_{0}^{\frac{\pi}{3}} (\sin 2x - \sin x)\,\mathrm{d}x\) (\(\sin 2x\) is the upper curve here)
M1 for integrating: \(\left[-\frac{1}{2}\cos 2x + \cos x\right]\)
A1 for \(\frac{1}{4}\)
Worked solution: For \(0 \lt x \lt \frac{\pi}{3}\), \(\sin 2x \gt \sin x\). Area \(= \left[-\frac{1}{2}\cos 2x + \cos x\right]_{0}^{\frac{\pi}{3}} = \left(\frac{1}{4} + \frac{1}{2}\right) - \left(-\frac{1}{2} + 1\right) = \frac{1}{4}\).
(c)Answer: \(\frac{5}{2}\)
B1 for the curves cross at \(x = \frac{\pi}{3}\); for \(\frac{\pi}{3} \lt x \lt \pi\), \(\sin x \gt \sin 2x\), so part of the integral is negative and cancels part of the area
M1 for \(\int_{\frac{\pi}{3}}^{\pi} (\sin x - \sin 2x)\,\mathrm{d}x\) evaluated
A1 for \(\frac{9}{4}\)
A1 for total area \(\frac{1}{4} + \frac{9}{4} = \frac{5}{2}\)
Worked solution: The integrand \(\sin 2x - \sin x\) is positive on \(\left(0, \frac{\pi}{3}\right)\) but negative on \(\left(\frac{\pi}{3}, \pi\right)\), so a single integral subtracts the second region's area from the first. \(\int_{\frac{\pi}{3}}^{\pi} (\sin x - \sin 2x)\,\mathrm{d}x = \left[-\cos x + \frac{1}{2}\cos 2x\right]_{\frac{\pi}{3}}^{\pi} = \left(1 + \frac{1}{2}\right) - \left(-\frac{1}{2} - \frac{1}{4}\right) = \frac{9}{4}\). Total area \(= \frac{1}{4} + \frac{9}{4} = \frac{5}{2}\).