Definite integration(10.1-10.3)

Recognise the concept of definite integration

  • Theory
    [TheoryMissing]
  • Examples
    Prove that \(\int_{\;h}^{\;k} {\,(x - h)(x - k)\,dx} = \frac{{{{(h - k)}^3}}}{6}\).
  • Solutions
    \(\begin{array}{1}\int_{\;h}^{\;k} {\,(x - h)(x - k)dx} = \int_{\;h}^{\;k} {\,[{x^2} - (h + k)x + hk]\,dx} \\
    = [\frac{{{x^3}}}{3} - \frac{{(h + k){x^2}}}{2} + hkx]{\kern 1pt} _h^k\\
    = (\frac{{{k^3}}}{3} - \frac{{h{k^2} + {k^3}}}{2} + h{k^2}) - (\frac{{{h^3}}}{3} - \frac{{{h^3} + {h^2}k}}{2} + {h^2}k)\\
    = \frac{{2{k^3} - 3h{k^2} - 3{k^3} + 6h{k^2} - 2{h^3} + 3{h^3} + 3{h^2}k - 6{h^2}k}}{6}\\
    = \frac{{{h^3} - 3{h^2}k + 3h{k^2} - {k^3}}}{6}\\
    = \frac{{{{(h - k)}^3}}}{6}\end{array}\)

Understand the properties of definite integrals

  • Theory
    [TheoryMissing]
  • Examples
    Assume that\(f(x)\) and \(g(x)\) are continuous functions, and \(\int_{\;2}^{\;4} {\;\frac{{f(x)}}{{f(x) - 2g(x)}}\,dx = 10} \). Evaluate the following definite integrals.
    (a) \(\int_{\;4}^{\;2} {\;\frac{{2f(x)}}{{f(x) - 2g(x)}}\,dx} \)
    (b) \(\int_{\;2}^{\;4} {\;\frac{{g(t)}}{{f(t) - 2g(t)}}\,dt} \)
  • Solutions
    (a)
    \(\begin{array}{1}\int_{\;4}^{\;2} {\;\frac{{2f(x)}}{{f(x) - 2g(x)}}\,dx} = - 2\int_{\;2}^{\;4} {\;\frac{{f(x)}}{{f(x) - 2g(x)}}\,dx}
    \\ = - 2{(10)^{^{}}}\\ = {\underline{\underline { - {\rm{ }}20}} ^{^{}}}\end{array}\)

    (b)
    \(\begin{array}{1}\int_{\;2}^{\;4} {\;\frac{{g(t)}}{{f(t) - 2g(t)}}\,dt} = \int_{\;2}^{\;4} {\;\frac{{g(x)}}{{f(x) - 2g(x)}}\,dx}
    \\ = - {\rm{ }}\frac{1}{2}{\int_{\;2}^{\;4} {\;\frac{{ - {\rm{ }}2g(x)}}{{f(x) - 2g(x)}}\,dx} ^{}}\\ = - {\rm{ }}\frac{1}{2}{\int_{\;2}^{\;4} {\;\frac{{[f(x) - 2g(x)] - f(x)}}{{f(x) - 2g(x)}}\,dx} ^{}}\\ = - {\rm{ }}\frac{1}{2}{\int_{\;2}^{\;4} {\,[1 - \frac{{f(x)}}{{f(x) - 2g(x)}}]\,dx} ^{}}\\ = - {\rm{ }}\frac{1}{2}\int_{\;2}^{\;4} {\,dx + } \frac{1}{2}{\int_{\;2}^{\;4} {\;\frac{{f(x)}}{{f(x) - 2g(x)}}\,dx} ^{}}\\ = - {\rm{ }}\frac{1}{2}[x]_{{\kern 1pt} 2}^{{\kern 1pt} 4} + \frac{1}{2}(10)\\ = - {\rm{ }}\frac{1}{2}(4 - 2) + {5^{^{^{}}}}\\ = {\underline{\underline 4} ^{}}\end{array}\)

Find definite integrals of algebraic functions, trigonometric functions and exponential functions

  • Theory
    [TheoryMissing]
  • Examples
    (a) Prove, by mathematical induction, that for all positive integers n,
    \(\sin x + \sin 2x + \sin 3x + \cdots + \sin nx = \frac{{\cos {\textstyle{x \over 2}} - \cos (n + {\textstyle{1 \over 2}})x}}{{2\sin {\textstyle{x \over 2}}}}\).
    (b) Find \(\mathop {\lim }\limits_{n \to \infty } n\sin \frac{\pi }{{2n}}\).
    (c) Hence, using the right end-point as \(x{\kern 1pt} _i^*\) of each subinterval \([{x_{i - 1}}{\rm{ }},{\rm{ }}{x_i}]\) of x, evaluate the definite integral \(\int_{\;0}^{\;\pi } {\,\sin x\,dx} \).
  • Solutions
    (a) Let \(P(n)\) be ‘\(\sin x + \sin 2x + \sin 3x + \cdots + \sin nx = \frac{{\cos {\textstyle{x \over 2}} - \cos (n + {\textstyle{1 \over 2}})x}}{{2\sin {\textstyle{x \over 2}}}}\)’.
    When \(n = 1\),
    L.H.S.\( = \sin x\)
    R.H.S.\(\begin{array}{l} = \frac{{\cos {\textstyle{x \over 2}} - \cos (1 + {\textstyle{1 \over 2}})x}}{{2\sin {\textstyle{x \over 2}}}}\\
    = \frac{{\cos {\textstyle{x \over 2}} - \cos {\textstyle{{3x} \over 2}}}}{{2\sin {\textstyle{x \over 2}}}}\\
    = \frac{{ - {\rm{ }}2\sin x\sin ( - {\textstyle{x \over 2}})}}{{2\sin {\textstyle{x \over 2}}}}\\
    = \frac{{ - {\rm{ }}\sin x( - \sin {\textstyle{x \over 2}})}}{{\sin {\textstyle{x \over 2}}}}\\
    = \sin x\end{array}\)
    ∴ \(P(1)\) is true.
    Assume that \(P(k)\) is true, where k is a positive integer.
    i.e. \(\sin x + \sin 2x + \sin 3x + \cdots + \sin kx = \frac{{\cos {\textstyle{x \over 2}} - \cos (k + {\textstyle{1 \over 2}})x}}{{2\sin {\textstyle{x \over 2}}}}\)
    then \(\begin{array}{l}\sin x + \sin 2x + \sin 3x + \cdots + \sin kx + \sin (k + 1)x\\
    = \frac{{\cos {\textstyle{x \over 2}} - \cos (k + {\textstyle{1 \over 2}})x}}{{2\sin {\textstyle{x \over 2}}}} + \sin (k + 1){x^{^{^{^{^{}}}}}}\\
    = \frac{{\cos {\textstyle{x \over 2}} - \cos (k + {\textstyle{1 \over 2}})x + 2\sin (k + 1)x\sin {\textstyle{x \over 2}}}}{{2\sin {\textstyle{x \over 2}}}}\\
    = \frac{{\cos {\textstyle{x \over 2}} - \cos (k + {\textstyle{1 \over 2}})x - [\cos (k + {\textstyle{3 \over 2}})x - \cos (k + {\textstyle{1 \over 2}})x]}}{{2\sin {\textstyle{x \over 2}}}}\\
    = \frac{{\cos {\textstyle{x \over 2}} - \cos \,[(k + 1) + {\textstyle{1 \over 2}}]\,x}}{{2\sin {\textstyle{x \over 2}}}}\end{array}\)
    ∴ \(P(k + 1)\)is true.
    According to mathematical induction, \(P(n)\) is true for all positive integers n.

    (b) Let \(t = \frac{\pi }{{2n}}\), then \(n = \frac{\pi }{{2t}}\).
    When \(n \to \infty \), \(t \to 0\).
    \(\begin{array}{1}\mathop {\lim }\limits_{n \to \infty } n\sin \frac{\pi }{{2n}} = \mathop {\lim }\limits_{t \to 0} \;\frac{\pi }{{2t}}\sin t\\
    = \frac{\pi }{2} \cdot \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}\\ = \frac{\pi }{2} \cdot 1\\
    = \underline{\underline {\frac{\pi }{2}}} \end{array}\)

    (c) Let \(f(x) = \sin x\).
    If [0,∏] is divided into n parts in equal width, then the width of each part is
    \(\Delta x = \frac{{b - a}}{n} = \frac{{\pi - 0}}{n} = \frac{\pi }{n}\)
    \(x{\kern 1pt} _i^*\, = a + i\Delta x = \frac{{i\pi }}{n}\)
    \(\begin{array}{1}f(x{\kern 1pt} _i^*{\kern 1pt} )\Delta x = \sin x{\kern 1pt} _i^*\, \cdot \frac{\pi }{n}\\
    = \frac{\pi }{n}\sin {\frac{{i\pi }}{n}^{}}\end{array}\)
    \(\begin{array}{1}\sum\limits_{i = 1}^n {f(x{\kern 1pt} _i^*{\kern 1pt} )\Delta x}
    = \sum\limits_{i = 1}^n {\frac{\pi }{n}\sin \frac{{i\pi }}{n}} \\
    = \frac{\pi }{n}{\sum\limits_{i = 1}^n {\sin i(\frac{\pi }{n})} ^{}}\\
    = \frac{\pi }{n} \cdot \frac{{\cos {\textstyle{\pi \over {2n}}} - \cos (n + {\textstyle{1 \over 2}}){\textstyle{\pi \over n}}}}{{2\sin {\textstyle{\pi \over {2n}}}}}\\
    = \frac{\pi }{n} \cdot {\frac{{\cos {\textstyle{\pi \over {2n}}} - \cos (\pi + {\textstyle{\pi \over {2n}}})}}{{2\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = \frac{\pi }{n} \cdot {\frac{{\cos {\textstyle{\pi \over {2n}}} + \cos {\textstyle{\pi \over {2n}}}}}{{2\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = {\frac{{2\pi \cos {\textstyle{\pi \over {2n}}}}}{{2n\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = {\frac{{\pi \cos {\textstyle{\pi \over {2n}}}}}{{n\sin {\textstyle{\pi \over {2n}}}}}^{}}\end{array}\)
    \(\begin{array}{1}\int_{\;0}^{\;\pi } {\,\sin x\,dx} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f(x{\kern 1pt} _i^*)\Delta x} \\
    = \mathop {\lim }\limits_{n \to \infty } \;{\frac{{\pi \cos {\textstyle{\pi \over {2n}}}}}{{n\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = {\frac{{\mathop {\lim }\limits_{n \to \infty } \pi \cos {\textstyle{\pi \over {2n}}}}}{{\mathop {\lim }\limits_{n \to \infty } n\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = {\frac{{\pi \cdot \mathop {\lim }\limits_{n \to \infty } \cos {\textstyle{\pi \over {2n}}}}}{{\mathop {\lim }\limits_{n \to \infty } n\sin {\textstyle{\pi \over {2n}}}}}^{}}\\
    = {\frac{{\pi \cdot 1}}{{{\textstyle{\pi \over 2}}}}^{}}\\
    = {\underline{\underline 2} ^{^{}}}\end{array}\)