Short and Sweet Calculus

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) area9; 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 \(a<b\).

The Riemann sum is an approximation to the net area if f takes on both positive and negative values.

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 \[\int_{3}^{-1}(2-x)dx=-4.\]


  1. or sometimes “net signed area”↩︎


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