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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