Applications of definite integration

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

  • Theory
  • Examples
    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.

    [GraphMissing M2E11 Q50]

    (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.
  • Solutions
    (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

  • Theory
  • Examples
    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.)
  • Solutions
    (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.