3.9 The Fundamental Theorem of Calculus

If \(f(x)\) is continuous on the interval \([a,b]\) and if \(F(x)\) is any indefinite integral of \(f(x)\) then \[\int ^b_a f(x) dx = F(b) - F(a)\]

Note.

If \(a = b\), then \(\displaystyle {\int ^b_a f(x) dx = 0}\).

Example 3.9.1. Evaluate the definite integrals

1.: \(\displaystyle {\int ^2_{-1}\dfrac {1}{x - 9}dx}\)
2.: \(\displaystyle {\int _0^{\frac {2\pi }{3}} \dfrac {d \theta }{5 + 4\cos \theta }}\)

Solution.

Part 1

\begin {align*} \int ^2_{-1}\dfrac {1}{x - 9}dx & = \ln \left |x - 9\right |\Bigg |^2_{-1}\\ & = \ln \left |2 - 9\right | - \ln \left |-1 - 9\right | = \ln 7 - \ln 10\\ & = \ln \dfrac {7}{10}\hspace {0.4cm} \text {or}\hspace {0.4cm} \ln (0.7)\\ \end {align*}

Part 2

Let \(\hspace {0.2cm} z = \tan \dfrac {\theta }{2}\hspace {0.2cm}\) then \(d \theta = \dfrac {2}{1 + z^2}dz\hspace {0.2cm}\) and \(\hspace {0.2cm} \cos \theta = \dfrac {1 - z^2}{1 + z^2}\hspace {0.2cm}\) Thus

\begin {align*} \int _0^{\dfrac {2\pi }{3}} \dfrac {d \theta }{5 + 4\cos \theta } & = \int ^{\sqrt {3}}_0 \frac {\dfrac {2}{1 + z^2}}{5 + 4\Bigg (\dfrac {1 - z^2}{1 + z^2}\Bigg )}\\\\ & = 2\int ^{\sqrt {3}}_0 \dfrac {1}{z^2 + 9}dz = \dfrac {2}{3}\tan ^{-1}\dfrac {z}{3}\Bigg |^{\sqrt {3}}_0\\\\ & = \dfrac {2}{3}\tan ^{\dfrac {\sqrt {3}}{3}} - \dfrac {2}{3}\tan ^{-1}\dfrac {0}{3}\\\\ & = \dfrac {2}{3}\Bigg (\dfrac {\pi }{6}\Bigg ) - \dfrac {2}{3}(0)\\ & = \dfrac {\pi }{9}\\\\ \end {align*}

Properties of Definite Integrals

If \(f(x)\) and \(g(x)\) are continuous functions on the interval of integration \([a,b]\), then

1.: \(\displaystyle {\int ^b_a K\hspace {0.1cm} f(x) dx = K\int ^b_a f(x) dx}\), \(\hspace {0.2cm} K = \) constant.
2.: \(\displaystyle {\int ^b_a \Big [ f(x)\pm g(x)\Big ] dx = \int ^b_a f(x) dx \pm \int ^b_a g(x) dx}\)
3.: For \(\hspace {0.2cm} a < c < b\), \(\hspace {0.3cm}\displaystyle {\int ^b_a f(x) dx = \int ^c_a f(x) dx + \int ^b_c f(x) dx}\)
4.: \(\displaystyle {\int ^b_a f(x) dx = - \int ^a_b f(x) dx}\)
5.: \(\displaystyle {\int ^b_a f(x) dx = (b- a) f(x_0)}\hspace {0.2cm}\) for at least one value \(x = x_0\) between \(a\) and \(b\). This is the mean value theorem for integration.

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