Use the approximation by step functions to calculate integrals.
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We will now give an example of an integral that will be computed
according to the definition. This will turn out to be really exhausting
even for this quite simple example.
Example 1. Consider the function \(f:[0,1]\to\mathbb{R}\) with \(f(x)=x\). Determine \(\int_0^1f(x)\, dx=\int_0^1x\, dx\). First
we consider two sequences of step functions \((\phi_n)_{n\in\mathbb{N}}\), \((\psi_n)_{n\in\mathbb{N}}\) with \[\begin{aligned}
\phi_n(x)&=\frac{k-1}n\text{ for
}x\in\left[\frac{k-1}n,\frac{k}n\right),\qquad k\in\{1,\ldots,n\},\\
\psi_n(x)&=\frac{k}n\text{ for
}x\in\left[\frac{k-1}n,\frac{k}n\right),\qquad k\in\{1,\ldots,n\}.
\end{aligned}\] Then for all \(n\in\mathbb{N}\) holds \(\phi_n\leq f\leq \psi_n\). Now we calculate
\[\begin{aligned}
\int_0^1\phi_n(x)\,
dx&=\sum_{k=1}^n\frac{k-1}n\cdot\left(\frac{k}n-\frac{k-1}n\right)\\
&=
\sum_{k=1}^n\frac{k-1}{n}\cdot\frac{1}n\\&=\frac{1}{n^2}\sum_{k=1}^n(k-1)\\&=\frac{1}{n^2}\cdot\frac{n(n-1)}2=\frac12-\frac{1}{2n}
\end{aligned}\] and \[\begin{aligned}
\int_0^1\psi_n(x)\,
dx&=\sum_{k=1}^n\frac{k}n\cdot\left(\frac{k}n-\frac{k-1}n\right)\\&=
\sum_{k=1}^n\frac{k}{n}\cdot\frac{1}n\\&=\frac{1}{n^2}\sum_{k=1}^nk\\&=\frac{1}{n^2}\cdot\frac{n(n+1)}2=\frac12+\frac{1}{2n}.
\end{aligned}\] In particular, we have for all \(n\in\mathbb{N}\) that \[\frac12-\frac{1}{2n}=\int_0^1\phi_n(x)\,
dx\leq\int_0^1x\, dx\leq \int_0^1\psi_n(x)\,
dx=\frac12+\frac{1}{2n}\] and thus \[\int_0^1x\, dx=\frac12.\]
By the definition of the integral, it is not difficult to obtain that
for \(f\in\mathcal{R}([a,b])\) and
\(c\in(a,b)\) holds \[\int_a^bf(x)\, dx=\int_a^cf(x)\,
dx+\int_c^bf(x)\, dx.\] To make this formula also valid for \(c\geq b\) or \(c\leq a\), we define for \(a\geq b\) that \[\int_a^bf(x)\, dx:=-\int_b^af(x)\,
dx.\]
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