Example 1.2.5. Revisiting Example 1.1.14.

Back in Example 1.1.14 we saw that when \(b \gt 0\) \(\int_0^b x\dee{x} =\frac{b^2}{2}\text{.}\) We’ll now verify that \(\int_0^b x\dee{x} =\frac{b^2}{2}\) is still true when \(b=0\) and also when \(b \lt 0\text{.}\)
  • First consider \(b=0\text{.}\) Then the statement \(\int_0^b x\dee{x} =\frac{b^2}{2}\) becomes
    \begin{gather*} \int_0^0 x\dee{x} =0 \end{gather*}
    This is an immediate consequence of Theorem 1.2.3(a).
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  • Now consider \(b \lt 0\text{.}\) Let us write \(B=-b\text{,}\) so that \(B \gt 0\text{.}\) In Example 1.1.14 we saw that
    \begin{gather*} \int_{-B}^0 x\dee{x} =-\frac{B^2}{2}. \end{gather*}
    So we have
    \begin{align*} \int_0^b x\dee{x} &=\int^{-B}_0 x\dee{x} =- \int_{-B}^0 x\dee{x} & \text{by Theorem }\knowl{./knowl/xref/thm_Intdomain.html}{\text{1.2.3}}\text{(b)}\\ & =-\left(-\frac{B^2}{2}\right) & \text{by Example }\knowl{./knowl/xref/eg_INTtriangle.html}{\text{1.1.14}}\\ & =\frac{B^2}{2} = \frac{b^2}{2} \end{align*}
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We have now shown that
\begin{align*} \int_0^b x\dee{x} &=\frac{b^2}{2} &\text{ for all real numbers $b$} \end{align*}
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