Short and Sweet Calculus

6.1Definition 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 $$a<b$$.

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”↩︎