7.1

Given the integral \(\int_5^{11}\frac8x\,\mathrm{d}x\).

Use the trapezium rule, with \(3\) strips each of width \(2\), to estimate the value of the integral. Give your answer correct to \(3\) significant figures.

\(\int_5^{11}\frac8x\,\mathrm{d}x \approx \frac{1}{2\times2} (\frac85+2\frac87+2\frac89+\frac8{11})\approx 1.598\)

7.2

The gradient of the curve given by \(\frac{dy}{dx}=x^2-2x+5\). The curve passes through the point \((3,11)\).

Find the equation of the curve.

\(y=\int(x^2-2x+5)\,\mathrm dx = \frac13x^3-x^2+5x+c\). Since the curve passes through the point \((3,11)\), we obtain:

\(\frac{3^3}3-3^2+5\times3+c=11\\ 15+c=11\\ c=-4\)

7.3

The diagram shows parts of the curves \(y=x^2+1\) and \(y=11-\frac{9}{x^2}\) , which intersect at \((1,2)\) and \((3, 10)\).

Use integration to find the exact area of the shaded region enclosed between the two curves.

The area under the curve \(y=11-\frac{9}{x^2}\) for \(1\leqslant x \leqslant 3\)

 is \(\int_1^3(11-\frac{9}{x^2})\,\mathrm{d}x = (11x + \frac{9}{x})\Big|_1^3 = 33+3-11-9=16\) and under the curve \(y=x^2+1\) for \(1\leqslant x \leqslant 3\) is \(\int_1^3(x^2+1)\,\mathrm{d}x = ({x^3 \over 3} + x)\Big|_1^3 = 9+3-\frac13-1=\frac{32}3\). Therefore, the area of the shaded region is \(16-\frac{32}3 = \frac{16}3\).

7.4

Find the integral \(\int x(x^2-4)\,\mathrm dx\).

Evaluate \(\int_0^2 x(x^2-4)\,\mathrm dx\).

\(\int_0^2 x(x^2-4)\,\mathrm dx = \int_0^2 (x^3-4x)\,\mathrm dx = \left.(\frac{x^4}{4}-\frac{4x^2}2)\right|_0^2 = \frac{16}{4}-\frac{4\times4}2 = -4\)

7.5

The diagram shows the curve \(y = \sqrt{4x + 1}\).

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 \( y = \sqrt{4x + 1}\), the \(x\text{-axis}\), and the lines \(x = 1\) and \(x = 3\). Give your answer correct to \(3\) significant figures.

The area is approximately equal to \(\int_1^{3}\sqrt{4x+1}\,\mathrm{d}x \approx \frac{2}{2\times4}(\sqrt5 + 2\sqrt{7}+2\sqrt{9}+2\sqrt{11}+\sqrt{13}) \approx 11.883\)

7.6

Given that \(\int_a^\infty \frac{24}{x^3}\, \mathrm dx = 3\).

Find the value of the positive constant \(a\).

\(\int_a^\infty \frac{24}{x^3}\, \mathrm dx = \int_a^\infty 24x^{-3}\, \mathrm dx = \left. \frac{24x^{-2}}{-2}\right|_a^\infty = 0 - (-12a^{-2}) = \frac{12}{a^2} = 3\), so

\(3a^2=12\\ a^2=4\)

Since \(a\) is positive, \(a=2\).

7.7

Find the integral \(\int (x^{3 \over 2} -1) \, \mathrm dx\).

Evaluate \(\int_1^4 (x^{3 \over 2} -1) \, \mathrm dx\).

\(\int_1^4 (x^{3 \over 2} -1) \, \mathrm dx = \left. \left( \frac{x^{5 \over 2}}{5\over 2} -x \right) \right|_1^4 = \frac{2\times 4^{5 \over 2}}{5} - 4 - \left(\frac{2\times 1^{5 \over 2}}{5} -1\right) = \frac{64}{5} - 3 -\frac25 = 12.4-3 = 9.4\)

7.8

The gradient of a curve is given by \(\frac{dy}{dx} = 6x-4\). The curve passes through the distinct points \((2, 5)\) and \((p, 5)\)

Find the value of \(p\).

The equation of the curve is \(y = \int (6x-4)\, \mathrm dx = {6x^2 \over 2} - 4x + c = 3x^2-4x + c\). Since it passes through the point \((2, 5)\), one can conclude that 

\(3\times2^2-4\times2 + c=5\\ 12-8+c=5\\ c=1\)

So, \(y = 3x^2-4x + 1\) and

\(3p^2-4p + 1 = 5\\ 3p^2-4p-4=0\\ D=4^2-4\times3\times(-4) = 16+48 = 64\\ p = \frac{4\pm\sqrt{64}}{6} = \frac{4\pm8}{6}\)

Therefore, \(p = 2\) or \(p = -\frac23\), but the given points are distinct, so \(p = -\frac23\).

7.9

Given the region, enclosed by the curve \(y = x^2+4x\), the \(x\text{-axis}\) and the lines \(x = 3\) and \(x = 6\).

Use integration to find the exact area of the region.

The area is equal to \(\int_3^6(x^2+4x)\, \mathrm dx = \left. (\frac{x^3}{3} + \frac{4x^2}{2}) \right|_3^6 = \frac{216}{3} + 2\times 36 - \frac{27}{3} - 2\times 9 = 72 + 72 - 9 - 18 = 117\)

7.10

Find the integral \(\int 4\sqrt x \, \mathrm dx\).

Evaluate \(\int_1^9 4\sqrt x \, \mathrm dx\).

\(\int_1^9 4\sqrt x \, \mathrm dx = \int_1^9 4x^{\frac12} \, \mathrm dx = \left. \frac{4x^{\frac32}}{3\over2} \right|_1^9 = \frac{8\times9^{\frac32}}{3} - \frac{8\times1^{\frac32}}{3} = \frac{8\times 27}{3} - \frac83 = 72-\frac83 = 69\frac13\)