Definite integration(10.4-10.6)

Use integration by substitution to find definite integrals

  • Theory
  • Examples
    Evaluate \(\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} \).
  • Solutions
    ∵ \(\frac{1}{{{{\cos }^2}( - x)\sqrt {4 - {{\tan }^2}( - x)} }} = \frac{1}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}\)
    ∴ \(\frac{1}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}\) is an even function.
    \(\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,\frac{{{{\sec }^2}x\,dx}}{{\sqrt {4 - {{\tan }^2}x} }}} \)
    Let \(u = \tan x\), then \(du = {\sec ^2}x\,dx\).
    When \(x = 0\), \(u = 0\).
    When \(x = \frac{\pi }{3}\), \(u = \sqrt 3 \).
    \(\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\sqrt 3 } {\,\frac{{du}}{{\sqrt {4 - {u^2}} }}} \)
    Let \(u = 2\sin \theta \), where \( - {\rm{ }}\frac{\pi }{2} < \theta < \frac{\pi }{{\rm{2}}}\),
    then \(du = {\rm{2}}\cos \theta \,d{\kern 1pt} \theta \) and \(\sqrt {4 - {u^2}} = 2\cos \theta \).
    When \(u = 0\), \(\theta = 0\).
    When \(u = \sqrt 3 \), \(\theta = \frac{\pi }{3}\).
    \(\begin{array}{1}\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,\frac{{2\cos \theta \,d{\kern 1pt} \theta }}{{2\cos \theta }}} \\
    = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,d\theta } \\
    = 2\,[\theta ]{\kern 1pt} _0^{\frac{\pi }{3}}\\
    = {\underline{\underline {\frac{{2\pi }}{3}}} ^{^{}}}\end{array}\)

Use integration by parts to find definite integrals

  • Theory
  • Examples
    Evaluate \(\int_{\;0}^{\;3} {\,{{(\frac{x}{{{e^{2x}}}})}^2}dx} \).
  • Solutions
    \(\begin{array}{1}\int_{\;0}^{\;3} {\,{{(\frac{x}{{{e^{2x}}}})}^2}dx} = \int_{\;0}^{\;3} {\,{x^2}{e^{ - 4x}}\,dx} \\
    = - {\rm{ }}\frac{1}{4}\int_{\;0}^{\;3} {\,{x^2}\,d({e^{ - 4x}})} \\
    = - {\rm{ }}\frac{1}{4}[{x^2}{e^{ - 4x}}]{\kern 1pt} _0^3\, + \frac{1}{4}\int_{\;0}^{\;3} {\,{e^{ - 4x}}\,d({x^2})} \\
    = - {\rm{ }}\frac{1}{4}(9{e^{ - 12}} - 0) + \frac{1}{2}\int_{\;0}^{\;3} {\,x{e^{ - 4x}}\,dx} \\
    = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}\int_{\;0}^{\;3} {\,xd({e^{ - 4x}})} \\
    = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}[x{e^{ - 4x}}]{\kern 1pt} _0^3\, + \frac{1}{8}\int_{\;0}^{\;3} {\,{e^{ - 4x}}\,dx} \\
    = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}(3{e^{ - 12}} - 0) - \frac{1}{{32}}[{e^{ - 4x}}]{\kern 1pt} _0^3\\
    = \underline{\underline { - {\rm{ }}\frac{{85}}{{32}}{e^{ - 12}} + \frac{1}{{32}}}} \end{array}\)

Understand the properties of the definite integrals of even, odd and periodic functions

  • Theory
    [TheoryMissing]
  • Examples
    Let \(f(x)\), \(g(x)\) and \(h(x)\) be odd functions, \(p(x)\) and \(q(x)\) be even functions.
    (a) Prove that the following functions are odd functions.
    (i) \(h(g(f(x)))\) (ii) \(f(g(x)) \cdot p(q(x))\)
    (b) Evaluate the following definite integrals.
    (i) \(\int_{\; - \,\frac{\pi }{2}}^{\;\frac{\pi }{2}} {\;{{\sin }^3}(2{x^5})\,dx} \) (ii) \(\int_{\; - \pi }^{\;\pi } {\,\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}\,dx} \)
  • Solutions
    (a) (i)
    ∵ \(\begin{array}{1}h{\kern 1pt} (g(f( - x))) = h{\kern 1pt} (g( - f(x)))\\ = h{\kern 1pt} ( - {\rm{ }}g(f(x)))\\
    = - {\rm{ }}h{\kern 1pt} (g(f(x)))\end{array}\)
    ∴ \(h{\kern 1pt} (g(f(x)))\) is an odd function.
    (ii)
    ∵ \(\begin{array}{1}f(g( - x)) \cdot p(q( - x)) = f( - g(x)) \cdot p(q(x))\\
    = - f(g(x)) \cdot p(q(x))\end{array}\)
    ∴ \(f(g(x)) \cdot p(q(x))\) is an odd function.

    (b) (i)
    ∵ \(\begin{array}{1}2{( - x)^5} = - {\rm{ }}2{x^5}\
    \\sin ( - x) = - {\rm{ }}\sin x
    \\{( - x)^3} = - {x^3}\end{array}\)
    ∴ Take \(f(x) = 2{x^5}\), \(g(x) = \sin x\) and \(h(x) = {x^3}\),
    \(\begin{array}{1}h(g(f(x))) = {[\sin (2{x^5})]^3}\\
    = {\sin ^3}(2{x^5})\end{array}\)
    ∴ \({\sin ^3}(2{x^5})\) is an odd function.
    Hence \(\int_{\; - \,\frac{\pi }{2}}^{\;\frac{\pi }{2}} {\;{{\sin }^3}(2{x^5})\,dx} = \underline{\underline 0} \).
    (ii)
    ∵ \(\begin{array}{1}\sqrt[3]{{ - {\rm{ }}x}} = - {\rm{ }}\sqrt[3]{x}\\
    - {\rm{ }}x - \sin ( - x) = - {\rm{ }}{(x - \sin x)^{}}\\
    \cos ( - x) = \cos {x^{}}\\
    {( - x)^2} = {x^2}\end{array}\)
    ∴ Take \(f(x) = \sqrt[3]{x}\), \(g(x) = x - \sin x\), \(p(x) = \cos x\) and \(q(x) = {x^2}\),
    \(\begin{array}{1}f(g(x)) \cdot p(q(x)) = \sqrt[3]{{x - \sin x}} \cdot \cos ({x^2})\\
    = \cos {x^2} \cdot {\sqrt[3]{{x - \sin x}}^{}}\end{array}\)
    ∴ \(\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}\) is an odd function.
    Hence \(\int_{\; - \pi }^{\;\pi } {\,\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}dx} = \underline{\underline 0} \).