Unit 28 Review Exercises
Applications of Integration
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 Evaluate \(\displaystyle\int_1^2 \left(6x^2 - \frac{2}{x^2}\right)\mathrm{d}x\).
\(\displaystyle\int (6x^2 - 2x^{-2})\,\mathrm{d}x = 2x^3 + \frac{2}{x}\).
\(\Big[2x^3 + \tfrac{2}{x}\Big]_1^2 = (16 + 1) - (2 + 2) = 13\). See 28.1.
2 Paper 1 Given that \(\displaystyle\int_0^3 f(x)\,\mathrm{d}x = 5\), find \(\displaystyle\int_0^3 \big(2f(x) - x\big)\,\mathrm{d}x\) and \(\displaystyle\int_3^0 f(x)\,\mathrm{d}x\).
\(2 \times 5 - \Big[\tfrac{x^2}{2}\Big]_0^3 = 10 - \tfrac{9}{2} = \dfrac{11}{2}\).
\(\displaystyle\int_3^0 f(x)\,\mathrm{d}x = -5\). See 28.1.
3 Paper 1 The region under \(y = x^2 + 1\) from \(x = 1\) to \(x = 3\) is divided into 4 strips of width \(0.5\).
(a) Find the lower sum and the upper sum.
(b) Find the exact area of the region.
Heights: \(f(1) = 2\), \(f(1.5) = 3.25\), \(f(2) = 5\), \(f(2.5) = 7.25\), \(f(3) = 10\).
(a) Lower: \(0.5\,(2 + 3.25 + 5 + 7.25) = 8.75\). Upper: \(0.5\,(3.25 + 5 + 7.25 + 10) = 12.75\).
(b) \(\Big[\tfrac{x^3}{3} + x\Big]_1^3 = (9 + 3) - \left(\tfrac{1}{3} + 1\right) = \dfrac{32}{3} \approx 10.7\), between the two sums. See 28.2.
4 Paper 1 Find the area of the region enclosed by \(y = x^2 - 6x + 5\) and the \(x\)-axis.
\((x - 1)(x - 5) = 0\): \(x = 1, 5\). The curve is below the axis between them.
\(\displaystyle\int_1^5 (x^2 - 6x + 5)\,\mathrm{d}x = \Big[\tfrac{x^3}{3} - 3x^2 + 5x\Big]_1^5 = \left(\tfrac{125}{3} - 50\right) - \tfrac{7}{3} = -\tfrac{32}{3}\).
Area \(= \dfrac{32}{3}\). See 28.3.
5 Paper 1 (a) Find \(\displaystyle\int_0^{\pi} \sin 2x\,\mathrm{d}x\).
(b) Find the total area between \(y = \sin 2x\) and the \(x\)-axis for \(0 \leq x \leq \pi\). Explain why the answers to (a) and (b) are different.
(a) \(\Big[-\tfrac{1}{2}\cos 2x\Big]_0^{\pi} = -\tfrac{1}{2} + \tfrac{1}{2} = 0\).
(b) \(\displaystyle\int_0^{\frac{\pi}{2}} \sin 2x\,\mathrm{d}x = \Big[-\tfrac{1}{2}\cos 2x\Big]_0^{\frac{\pi}{2}} = \tfrac{1}{2} + \tfrac{1}{2} = 1\). By symmetry the part from \(\frac{\pi}{2}\) to \(\pi\) is below the axis with area \(1\). Total \(= 2\).
In (a) the part below the axis counts as negative and cancels the part above. See 28.3.
6 Paper 1 Find the area enclosed by the curves \(y = x^2\) and \(y = 4x - x^2\).
\(x^2 = 4x - x^2 \Rightarrow 2x^2 - 4x = 0 \Rightarrow x = 0, 2\). The curve \(y = 4x - x^2\) is above.
\(\displaystyle\int_0^2 (4x - 2x^2)\,\mathrm{d}x = \Big[2x^2 - \tfrac{2x^3}{3}\Big]_0^2 = 8 - \tfrac{16}{3} = \dfrac{8}{3}\). See 28.4.
7 Paper 2 The curves \(y = \mathrm{e}^{x/2}\) and \(y = 4 - x^2\) meet at two points.
(a) Find the \(x\)-coordinates of the two points.
(b) Find the area of the region enclosed by the curves.
(a) GDC: \(x = -1.9009\ldots \approx -1.90\) and \(x = 1.4068\ldots \approx 1.41\).
(b) \(y = 4 - x^2\) is above (at \(x = 0\): \(4 > 1\)). \(\displaystyle\int_{-1.9009\ldots}^{1.4068\ldots} \left(4 - x^2 - \mathrm{e}^{x/2}\right)\mathrm{d}x = 6.745\ldots \approx 6.75\). See 28.4.
8 Paper 1 AHL Find the area of the region enclosed by the curve \(x = y^2 - 2y\) and the \(y\)-axis.
\(y^2 - 2y = 0 \Rightarrow y = 0, 2\). For \(0 < y < 2\), \(x < 0\): the curve is to the left of the \(y\)-axis.
\(\displaystyle\int_0^2 (y^2 - 2y)\,\mathrm{d}y = \tfrac{8}{3} - 4 = -\tfrac{4}{3}\), so the area is \(\dfrac{4}{3}\). See 28.3.
9 Paper 1 AHL The region enclosed by \(y = 2\sqrt{x}\) and \(y = x\) is rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid.
\(2\sqrt{x} = x \Rightarrow x = 0\) or \(x = 4\). \(y = 2\sqrt{x}\) is above (at \(x = 1\): \(2 > 1\)).
\(V = \pi\displaystyle\int_0^4 \left(4x - x^2\right)\mathrm{d}x = \pi\left(32 - \tfrac{64}{3}\right) = \dfrac{32\pi}{3}\). See 28.5.
10 Paper 1 AHL The region bounded by the curve \(y = \ln(x + 1)\), the \(y\)-axis and the line \(y = 2\) is rotated through \(2\pi\) about the \(y\)-axis. Show that the volume of the solid is \(\dfrac{\pi}{2}\left(\mathrm{e}^4 - 4\mathrm{e}^2 + 7\right)\).
\(y = \ln(x + 1) \Rightarrow x = \mathrm{e}^y - 1\). The curve passes through the origin, so \(y\) goes from \(0\) to \(2\).
\(V = \pi\displaystyle\int_0^2 (\mathrm{e}^y - 1)^2\,\mathrm{d}y = \pi\int_0^2 \left(\mathrm{e}^{2y} - 2\mathrm{e}^y + 1\right)\mathrm{d}y = \pi\Big[\tfrac{1}{2}\mathrm{e}^{2y} - 2\mathrm{e}^y + y\Big]_0^2\)
\(= \pi\left[\left(\tfrac{\mathrm{e}^4}{2} - 2\mathrm{e}^2 + 2\right) - \left(\tfrac{1}{2} - 2\right)\right] = \pi\left(\tfrac{\mathrm{e}^4}{2} - 2\mathrm{e}^2 + \tfrac{7}{2}\right) = \dfrac{\pi}{2}\left(\mathrm{e}^4 - 4\mathrm{e}^2 + 7\right)\). See 28.5.