###### 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 $$\int_1^9 4\sqrt x \, \mathrm dx$$.

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