### Applications of definite integration

#### Understand the application of definite integrals in finding the area of a plane figure

• In the figure, the curves $${C_{{\rm{ }}1}}{\kern 1pt} :y = {x^2}{\rm{ }}(x > 0)$$ and $${C_{{\rm{ }}2}}{\kern 1pt} :y = \frac{{24}}{x} + 1{\rm{ }}(x > 0)$$ intersect at Q. L is the tangent to $${C_{{\rm{ }}1}}$$ at P(1, 1). L and $${C_{{\rm{ }}2}}$$ intersect at R.

(a) Find the equation of L.
(b) Find the coordinates of Q and R.
(c) Find the area bounded by $${C_{{\rm{ }}1}}$$, $${C_{{\rm{ }}2}}$$ and L.
• (a)
$$\begin{array}{1}y = {x^2}\\ \frac{{dy}}{{dx}} = 2x\\ {\left. {\frac{{dy}}{{dx}}{\kern 1pt} } \right|_{{\kern 1pt} x = 1}}{\kern 1pt} = 2(1)\\ = 2\end{array}$$
Slope of L$$= 2$$
The equation of L is
$$\begin{array}{1}y - 1 = 2(x - 1)\\ 2x - y - 1 = 0\end{array}$$

(b) $$\left\{ \begin{array}{l}y = {x^2}\;......................\;(1)\\y = \frac{{24}}{x} + 1\;................\;(2)\end{array} \right.$$
Substitute (1) into (2),
$$\begin{array}{1}{x^2} = \frac{{24}}{x} + 1 \\{x^3} - x - 24 = 0 \\(x - 3)({x^2} + 3x + 8) = 0 \\x = 3\end{array}$$
Substitute $$x = 3$$ into (1),
$$\begin{array}{1}y = {3^2}\\ = 9\end{array}$$
∴ The coordinates of Q are (3, 9).
$$\left\{ \begin{array}{l}y = \frac{{24}}{x} + 1\;.....................\;(2)\\2x - y - 1 = 0\;................\;(3)\end{array} \right.$$
Substitute (2) into (3),
$$\begin{array}{1}2x - (\frac{{24}}{x} + 1) - 1 = 0 \\2x - \frac{{24}}{x} - 2 = 0 \\{x^2} - x - 12 = 0\\(x - 4)(x + 3) = 0 \\x = 4\;\;{\rm{o}}{\kern 1pt} {\rm{r}}\;\;x = - 3\;({\rm{r ej e c t e d}})\end{array}$$
Substitute $$x = 4$$ into (2),
$$\begin{array}{1}y = \frac{{24}}{4} + 1\\ = 7\end{array}$$
∴ The coordinates of R are (4, 7).

(c) Rewrite the equation $$2x - y - 1 = 0$$ as $$y = 2x - 1$$.
Required area$$\begin{array}{l} = \int_{\;1}^{\;3} {\;[{x^2} - (2x - 1)]\,dx} + \int_{\;3}^{\;4} {\;[(\frac{{24}}{x} + 1) - (2x - 1)]\,dx} \\ = \int_{\;1}^{\;3} {\;({x^2} - 2x + 1)\,dx} + \int_{\;3}^{\;4} {\;(\frac{{24}}{x} - 2x + 2)\,dx} \\ = [\frac{{{x^3}}}{3} - {x^2} + x]{\kern 1pt} _1^3\, + [24\ln {\kern 1pt} \left| {{\kern 1pt} x{\kern 1pt} } \right| - {x^2} + 2x]{\kern 1pt} _3^4\\ = \frac{8}{3} + 24\ln \frac{4}{3} - 5\\ = \underline{\underline {24\ln \frac{4}{3} - \frac{7}{3}}} \end{array}$$

#### Understand the application of definite integrals in finding the volume of a solid of revolution about a coordinate axis or a line parallel to a coordinate axis

• Figure I shows the curve , where . C is revolved about the y-axis to form a container (see Figure II). Assume that the volume of water is V cubic units when the depth of water inside the container is h units.
(a) Express V in terms of h.
(b) Find the capacity of the container.
(c) Initially, the container is empty. Water is now poured into the container at a rate of 16∏ cubic units/second. What is the rate of change of the depth of water when half of the container is filled? (Give your answer correct to 3 significant figures.)
• (a) When $$y = h$$,
$$\begin{array}{1}{(x - 8)^2} = h\\x = 8 \pm \sqrt h \end{array}$$
$$V = \int_{\;8 - \sqrt h }^{\;8 + \sqrt h } {\,2\pi x\,[h - {{(x - 8)}^2}]\,dx}$$
Let $$u = x - 8$$, $$du = dx$$.
When $$x = 8 - \sqrt h$$, $$u = - {\rm{ }}\sqrt h$$.
When $$x = 8 + \sqrt h$$, $$u = \sqrt h$$.
$$\begin{array}{1}\therefore V=\int^{\sqrt{h}}_{-\sqrt{h}}{2\pi \left(u+8\right)\left(h-u^2\right)du}\\ =2\pi \int^{\sqrt{h}}_{-\sqrt{h}}{\left(-u^3-8u^2+hu+8h\right)du}\\ =2\pi {\left[-\frac{u^4}{4}-\frac{8u^3}{3}+\frac{hu^2}{2}+8hu\right]}^{\sqrt{h}}_{-\sqrt{h}}\\ =\frac{64\pi h\sqrt{h}}{3}\ \end{array}$$

(b) When $$x = 10$$,
$$\begin{array}{1}y = {(10 - 8)^2}\\ = 4\end{array}$$
∴ Capacity $$= \frac{{64\pi (4)\sqrt 4 }}{3}$$ cubic units
$$= \frac{{512\pi }}{3}$$ cubic units

(c) When half of the container is filled,
$$\begin{array}{c}V = \frac{{512\pi }}{3} \times \frac{1}{2}\\ = {\frac{{256\pi }}{3}^{}}\end{array}$$
$$\begin{array}{c}\frac{{64\pi h\sqrt h }}{3} = \frac{{256\pi }}{3}\\ {h^{\frac{3}{2}}} = 4\\h = {4^{\frac{2}{3}}}\end{array}$$
$$\begin{array}{c}V = \frac{{64\pi {h^{\frac{3}{2}}}}}{3}\\ \frac{{dV}}{{dt}} = \frac{{64\pi }}{3} \cdot \frac{3}{2}{h^{\frac{1}{2}}} \cdot \frac{{dh}}{{dt}}\\ = 32\pi {h^{\frac{1}{2}}}{\frac{{dh}}{{dt}}^{^{^{}}}}\end{array}$$
∵ $$\frac{{dV}}{{dt}} = 16\pi$$ and $$h = {4^{\frac{2}{3}}}$$
∴ $$\begin{array}{c}16\pi = 32\pi {\kern 1pt} {({4^{\frac{2}{3}}})^{\frac{1}{2}}}\frac{{dh}}{{dt}}\\\frac{{dh}}{{dt}} = \frac{1}{{{2^{\frac{5}{3}}}}}\end{array}$$
$$= 0.315$$ (corr. to 3 sig. fig.)
∴ The rate of change of the depth of water is 0.315 unit/second.