2.3: Properties of Definite Integrals

Suppose \(f\) is a continuous function on a closed interval \([a, b] .\) Let \(N\) be a positive infinite integer, \(d x=\frac{b-a}{N},\) and, for \(i=1,2, \cdots, N,\) let \(x_{i}^{*}\) a number in the \(i\)th subinterval of \([a, b]\) when it is partitioned into \(N\) intervals of equal length \(d x .\)

We first note that if \(f(x)=1\) for all \(x\) in \([a, b],\) then

\[\sum_{i=1}^{N} f\left(x_{i}^{*}\right) d x=\sum_{i=1}^{N} d x=b-a\] since the sum of the lengths of the subintervals must be the length of the interval. Hence

\[\int_{a}^{b} f(x) d x=\int_{a}^{b} d x=b-a .\] More generally, if \(f(x)=k\) for all \(x\) in \([a, b],\) where \(k\) is a fixed real number, then

\[\sum_{i=1}^{N} f\left(x_{i}^{*}\right) d x=\sum_{i=1}^{N} k d x=k \sum_{i=1}^{N} d x=k(b-a) ,\] and so

\[\int_{a}^{b} f(x) d x=\int_{a}^{b} k d x=k(b-a) .\] That is, the definite integral of a constant is the constant times the length of the interval. In particular, the integral of 1 over an interval is simply the lengthof the interval. If \(f\) is an arbitrary continuous function and \(k\) is a fixed constant, then

\[\sum_{i=1}^{N} k f\left(x_{i}^{*}\right) d x=k \sum_{i=1}^{N} f\left(x_{i} *\right) d x ,\] and so

\[\int_{a}^{b} k f(x) d x=k \int_{a}^{b} f(x) d x .\] That is, the definite integral of a constant times \(f\) is the constant times the definite integral of \(f .\) If \(g\) is also a continuous function on \([a, b],\) then

\[\sum_{i=1}^{N}\left(f\left(x_{i}^{*}\right)+g\left(x_{i}^{*}\right)\right) d x=\sum_{i=1}^{N} f\left(x_{i}^{*}\right) d x+\sum_{i=1}^{N} g\left(x_{i}^{*}\right) d x ,\] and so

\[\int_{a}^{b}(f(x)+g(x)) d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x .\] Now suppose \(c\) is another real number with \(a<c<b\). If the closed interval \([a, c]\) is divisible into \(M\) intervals of length \(d x,\) where \(M\) is a positive infinite integer less than \(N,\) then

\[\sum_{i=1}^{N} f\left(x_{i}^{*}\right) d x=\sum_{i=1}^{M} f\left(x_{i}^{*}\right) d x+\sum_{i=M+1}^{N} f\left(x_{i}^{*}\right) d x \] implies that

\[\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x .\] This is a reflection of our intuition that, for an object moving along a straight line, the change in position from time \(t=a\) to time \(t=b\) is equal to the change in position from time \(t=a\) time \(t=c\) plus the change in position from time \(t=c\) to time \(t=b\). Although we assumed that \([a, c]\) was divisible into an integer number of subintervals of length \(d x,\) the result holds in general. The final properties which we will consider revolve around a basic inequality. If \(f\) and \(g\) are both continuous on \([a, b]\) with \(f(x) \leq g(x)\) for all \(x\) in \([a, b],\) then

\[\sum_{i=1}^{N} f\left(x_{i}^{*}\right) d x \leq \sum_{i=1}^{N} g\left(x_{i}^{*}\right) d x ,\] from which it follows that

\[\int_{a}^{b} f(x) d x \leq \int_{a}^{b} g(x) d x .\] For example, if \(m\) and \(M\) are constants with \(m \leq f(x) \leq M\) for all \(x\) in \([a, b],\) then

\[\int_{a}^{b} m d x \leq \int_{a}^{b} f(x) d x \leq \int_{a}^{b} M d x ,\] and so

\[m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a) .\] Note, in particular, that if \(f(x) \geq 0\) for all \(x\) in \([a, b],\) then

\[\int_{a}^{b} f(x) d x \geq 0 .\]

Example \(\PageIndex{1}\)

From the observation that

\[f(x)=\frac{1}{1+x^{2}} \nonumber\]

is increasing on \((-\infty, 0]\) and decreasing on \([0, \infty),\) it is easy to see that

\[\frac{1}{2} \leq \frac{1}{1+x^{2}} \leq 1 \nonumber\]

for all \(x\) in \([-1,1] .\) Hence

\[1 \leq \int_{-1}^{1} \frac{1}{1+x^{2}} d x \leq 2 .\nonumber\]

We will eventually see, in Example \(2.6.20,\) that

\[\int_{-1}^{1} \frac{1}{1+x^{2}} d x=\frac{\pi}{2} \approx 1.5708 . \nonumber\]

Since for any real number \(a,-|a| \leq a \leq|a|\) (indeed, either \(a=|a|\) or \(a=-|a|),\) we have

\[-|f(x)| \leq f(x) \leq|f(x)| \] for all \(x\) in \([a, b] .\) Hence

\[-\int_{a}^{b}|f(x)| d x \leq \int_{a}^{b} f(x) d x \leq \int_{a}^{b}|f(x)| d x ,\nonumber\]

or, equivalently,

\[\left|\int_{a}^{b} f(x) d x\right| \leq \int_{a}^{b}|f(x)| d x . \nonumber\]

Notice that, since the definite integral is just a generalized version of summation, this result is a generalization of the triangle inequality: Given any real numbers \(a\) and \(b\),

\[|a+b| \leq|a|+|b| .\] The next theorem summarizes the properties of definite integrals that we have discussed above.

Theorem \(\PageIndex{1}\)

Suppose \(f\) and \(g\) are continuous functions on \([a, b], c\) is any real number with \(a<c<b,\) and \(k\) is a fixed real number. Then

1. \(\int_{a}^{b} k d x=k(b-a)\),
2. \(\int_{a}^{b} k f(x) d x=k \int_{a}^{b} f(x) d x\),
3. \(\int_{a}^{b}(f(x)+g(x)) d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x\),
4. \(\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\),
5. if \(f(x) \leq g(x)\) for all \(x\) in \([a, b],\) then \(\int_{a}^{b} f(x) d x \leq \int_{a}^{b} g(x) d x\),
6. if \(m \leq f(x) \leq M\) for all \(x\) in \([a, b],\) then \(m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)\),
7. \(\left|\int_{a}^{b} f(x) d x\right| \leq \int_{a}^{b}|f(x)| d x\).

Exercise \(\PageIndex{1}\)

Show that

\[\frac{1}{2} \leq \int_{1}^{2} \frac{1}{x} d x \leq 1 .\]