## 6.1 Definition of the Definite Integral

Suppose \(f(x)\geq0\) is a continuous function for \(a\leq x\leq b\). To calculate the area of the region bound by the curve \(y=f(x)\), the \(x\)-axis, and two vertical lines \(x=a\) and \(x=b\), we divide the area into \(n\) rectangles of equal width \(\Delta x=(b-a)/n\). The height of the \(i\)-th rectangle is \(f(x_{i}^{*})\) for some \(x_{i}^{*}\) within the base of the rectangle (Figure 6.1). The area is approximately equal to the sum of the areas of these rectangles: \[A\approx f(x_{1}^{*})\Delta x+f(x_{2}^{*})\Delta x+\cdots+f(x_{n}^{*})\Delta x.\tag{i}\] The right-hand side of (i) is called a Riemann sum. Using the sigma notation, we can write \[A\approx\sum_{i=1}^{n}f(x_{i}^{*})\Delta x.\]

If we increase the number of thin rectangles \(n\), thereby decreasing the width of the
rectangles \(\Delta x\), a better
approximation is obtained (Figure ???).
As \(n\to\infty\), the approximation
becomes equality; that is \[A=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\Delta
x.\] This limit is called the **definite integral**
of \(f(x)\) between \(a\) and \(b\) and is denoted by \({\int_{a}^{b}f(x)}dx\).

** 6.1**. \[\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\Delta
x\]

The use of the word “integral” and of the symbol \(\int\) suggests a connection with the indefinite integrals in the previous Chapter. This connection will be shown in Section [sec:Ch8-fundamental-theorem].

The symbol \(\int\) is a modified form of \(S\) (that stands for summation).

We call \(a\) the

**lower limit**and \(b\) the**upper limit**of integration (or of the internal), and the function \(f(x)\) the**integrand**.

If \(f(x)\) takes both positive and
negative values on \([a,b]\), then
\(\sum_{i=1}^{n}f(x_{i}^{*})\Delta x\)
is an approximation for the **net** (or
**signed**) **area**^{9};
that is, the area below the curve \(y=f(x)\) and above the \(x\)-axis minus the area above the curve
\(y=f(x)\) and below the \(x\)-axis from \(x=a\) to \(x=b\), as suggested in Figure 6.2. So we can say:

\({\displaystyle \int_{a}^{b}f(x)dx}=\) area above the \(x\)-axis \(-\) area below the \(x\)-axis

provided \(a<b\).

In the previous discussion, we assumed \(a<b\). However, even if \(b<a\), we can still retain our
definition of integral. If the upper and lower limits of integration are
reversed, \(\Delta x\) changes from
\((b-a)/n\) to \((a-b)/n\), so \[\sum_{i=1}^{n}f(x_{i}^{*})\left(\frac{b-a}{n}\right)=-\sum_{i=1}^{n}f(x_{i}^{*})\left(\frac{a-b}{n}\right).\]
We are thus led to the relation \[\boxed{\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx.}\]
If \(a=b\), then \(\Delta x=0\) and thus **\[\boxed{\int_{a}^{a}f(x)dx=0.}\]**

**Example 6.1**. Find \({\displaystyle \int_{-1}^{3}(2-x)dx}\).

**Solution**

Because the graph of \(y=2-x\) is a straight line, we can easily calculate the net signed area of the region lying between its graph and the \(x\)-axis. We should subtract the area of the part that lies below the \(x\)-axis from the area of the above that is above the \(x\)-axis (see Figure 6.3). It follows from the familiar triangle area formula \(A=\text{base}\times\text{height}/2\) that \[\int_{-1}^{3}(2-x)dx=\frac{3\times3}{2}-\frac{1\times1}{2}=4.\]

**Example 6.2**. Find \({\displaystyle \int_{3}^{-1}(2-x)dx}\).

**Solution**

Because \(\int_{3}^{-1}(2-x)dx=-\int_{-1}^{3}(2-x)dx\), and in the previous example, we showed that \(\int_{-1}^{3}(2-x)dx=4,\) we have \[\int_{3}^{-1}(2-x)dx=-4.\]

or sometimes “net signed area”↩︎

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