IGCSE · Practice questions · Definite integrals and areas

From limits to areas.

Six original Cambridge-style questions on definite integrals and plane areas: evaluating with exact values in ln, e and pi, regions above and below the \(x\)-axis, areas between a line and a curve and between two curves, and unknown limits.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch, find the limits, then integrate.

Integrate and write \(\big[F(x)\big]_{a}^{b}\); no \(+\,c\).
Substitute top limit minus bottom limit, bracketing the bottom value; keep \(\ln\), \(e\), \(\pi\) exact and use radians.
Find the limits of an area from the roots of \(y = 0\) or from solving the two equations simultaneously.
Split at roots: a region below the \(x\)-axis gives a negative integral, so add the magnitudes.
Between graphs integrate (top minus bottom), testing one \(x\)-value to see which is on top.

Calculation

[3 marks]

Evaluate \(\displaystyle\int_{1}^{4} \big(6\sqrt{x} - 2\big)\,dx\).

\(6\sqrt{x} = 6x^{1/2}\) integrates to \(\dfrac{6x^{3/2}}{3/2} = 4x^{3/2}\). ✓
\(\Big[4x^{3/2} - 2x\Big]_{1}^{4} = (4 \times 8 - 8) - (4 - 2)\). ✓
\(= 24 - 2 = 22\). ✓

Calculation

[5 marks]

Find the exact value of

(a) \(\displaystyle\int_{0}^{\pi/4} \big(\sec^{2} x + \sin 2x\big)\,dx\),

(b) \(\displaystyle\int_{0}^{1} \Big(e^{2x} + \frac{2}{x + 1}\Big)\,dx\).

(a) \(\Big[\tan x - \tfrac{1}{2}\cos 2x\Big]_{0}^{\pi/4} = \big(1 - \tfrac{1}{2}\cos\tfrac{\pi}{2}\big) - \big(0 - \tfrac{1}{2}\cos 0\big)\). ✓
\(= (1 - 0) - \big(-\tfrac{1}{2}\big) = \tfrac{3}{2}\). ✓
(b) \(\Big[\tfrac{1}{2}e^{2x} + 2\ln(x + 1)\Big]_{0}^{1} = \big(\tfrac{1}{2}e^{2} + 2\ln 2\big) - \big(\tfrac{1}{2} + 2\ln 1\big)\). ✓
\(= \tfrac{1}{2}e^{2} - \tfrac{1}{2} + 2\ln 2\) (or \(\tfrac{1}{2}(e^{2} - 1) + \ln 4\)). ✓

Analysis

[6 marks]

The curve \(y = x^{3} - 5x^{2} + 6x\) passes through the origin \(O\) and crosses the \(x\)-axis again at \(A\) and \(B\).

(a) Find the \(x\)-coordinates of \(A\) and \(B\).

(b) Find the total area of the two regions enclosed by the curve and the \(x\)-axis between \(O\) and \(B\).

(a) \(x(x^{2} - 5x + 6) = x(x - 2)(x - 3) = 0\), so \(A\) is at \(x = 2\) and \(B\) at \(x = 3\). ✓
(b) Let \(F(x) = \tfrac{1}{4}x^{4} - \tfrac{5}{3}x^{3} + 3x^{2}\). \(F(0) = 0\), \(F(2) = 4 - \tfrac{40}{3} + 12 = \tfrac{8}{3}\), \(F(3) = \tfrac{81}{4} - 45 + 27 = \tfrac{9}{4}\). ✓
From 0 to 2 (above the axis): \(\tfrac{8}{3}\). ✓
From 2 to 3 (below the axis): \(\tfrac{9}{4} - \tfrac{8}{3} = -\tfrac{5}{12}\), an area of \(\tfrac{5}{12}\). ✓
Total area \(= \tfrac{8}{3} + \tfrac{5}{12} = \tfrac{37}{12}\) square units. ✓

Calculation

[5 marks]

The line \(y = 2x + 5\) meets the curve \(y = 8 - x^{2}\) at the points \(P\) and \(Q\).

(a) Find the coordinates of \(P\) and \(Q\).

(b) Find the area of the region enclosed by the line and the curve.

(a) \(8 - x^{2} = 2x + 5 \Rightarrow x^{2} + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0\). ✓
\(P(-3, -1)\) and \(Q(1, 7)\). ✓
(b) At \(x = 0\) the curve gives 8 and the line 5, so the curve is on top: \(\displaystyle\int_{-3}^{1} \big(8 - x^{2} - (2x + 5)\big)\,dx = \int_{-3}^{1} (3 - 2x - x^{2})\,dx\). ✓
\(\Big[3x - x^{2} - \tfrac{1}{3}x^{3}\Big]_{-3}^{1} = \big(3 - 1 - \tfrac{1}{3}\big) - \big(-9 - 9 + 9\big) = \tfrac{5}{3} + 9\). ✓
Area \(= \tfrac{32}{3}\) square units. (Integrating top minus bottom handles the part of the region below the \(x\)-axis automatically.) ✓

Analysis

[5 marks]

The curves \(y = \sin x\) and \(y = \cos x\) meet at \(x = \dfrac{\pi}{4}\) and at \(x = \dfrac{5\pi}{4}\).

(a) Show that \(\sin x > \cos x\) at \(x = \dfrac{\pi}{2}\).

(b) Find the exact area of the region enclosed by the two curves between these points.

(a) At \(x = \tfrac{\pi}{2}\): \(\sin x = 1\) and \(\cos x = 0\), so \(y = \sin x\) is the upper curve. ✓
(b) Area \(= \displaystyle\int_{\pi/4}^{5\pi/4} (\sin x - \cos x)\,dx = \Big[-\cos x - \sin x\Big]_{\pi/4}^{5\pi/4}\). ✓
At \(\tfrac{5\pi}{4}\): \(-\big(-\tfrac{\sqrt{2}}{2}\big) - \big(-\tfrac{\sqrt{2}}{2}\big) = \sqrt{2}\). ✓
At \(\tfrac{\pi}{4}\): \(-\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2} = -\sqrt{2}\). ✓
Area \(= \sqrt{2} - (-\sqrt{2}) = 2\sqrt{2}\) square units. ✓

Analysis

[5 marks]

(a) Given that \(\displaystyle\int_{2}^{k} \frac{1}{x - 1}\,dx = \ln 4\), where \(k > 2\), find \(k\).

(b) The region bounded by \(y = e^{2x}\), the \(x\)-axis, the \(y\)-axis and the line \(x = k\), where \(k > 0\), has area 4. Find the exact value of \(k\).

(a) \(\Big[\ln(x - 1)\Big]_{2}^{k} = \ln(k - 1) - \ln 1 = \ln(k - 1)\). ✓
\(\ln(k - 1) = \ln 4 \Rightarrow k - 1 = 4 \Rightarrow k = 5\). ✓
(b) \(\Big[\tfrac{1}{2}e^{2x}\Big]_{0}^{k} = \tfrac{1}{2}e^{2k} - \tfrac{1}{2} = 4\). ✓
\(e^{2k} = 9 \Rightarrow 2k = \ln 9\). ✓
\(k = \tfrac{1}{2}\ln 9 = \ln 3\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.