In this section, we prove some key properties of definite integrals that will help us evaluate definite integrals in an easy way. Most of these properties are analogous to the properties of the summation process that we reviewed in Section 12.3.

1. $$\int_{a}^{b}[f(x)+g(x)]dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx}$$
2. $$\int_{a}^{b}cf(x)dx=c\int_{a}^{b}f(x)}dx$$ where $$c$$ is a constant.
In particular $$\int_{a}^{b}-f(x)dx=-\int_{a}^{b}f(x)dx}$$
3. $$\int_{a}^{b}cdx=c(b-a)$$, where $$c$$ is a constant.
4. $$\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\int_{a}^{b}f(x)dx}$$ either $$c$$ is between $$a$$ and $$b$$ or outside.
1. Property (1) says that the integral of a sum is the sum of the separate integrals (see Figure 1). The truth of this formula follows at once from the definition of the definite integral and the following property of the summation process $\sum_{i=1}^{n}(a_{i}+b_{i})=\sum_{i=1}^{n}a_{i}+\sum_{k=1}^{n}b_{i}.$ Figure 1. The figure shows $$\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g$$.
2. Property (2) says that we can move a constant factor outside the integral sign. Note that $$c$$ must be a constant (not a function of $$x$$). To prove, we use the definition of the definite integral and the following property: $\sum_{i=1}^{n}ca_{i}=c\sum_{i=1}^{n}a_{i},$
3. Property (3) says that the integral of a cosntant function $$f(x)=c$$ from $$a$$ to $$b$$ is $$c(b-a)$$. This is expected because if $$a<b$$, the signed area of the shaded rectangle in Figure 2 is $$c(b-a)$$. Figure 2. $$\int_{a}^{b}c\,dx}$$ signed area of the region between $$y=c$$ and the $$x$$-axis from $$x=a$$ to $$x=b$$ is $$c(b-a)$$.

Property (4) says to integrate a function $$f$$ from $$a$$ to $$b$$, we can integrate it once from $$a$$ to $$c$$, once from $$c$$ to $$b$$, and then add the results together.

If $$a<c<b$$, the truth of property (4) can be easily shown graphically (see Figure 3). Also, it is easy to see from the definition of the definite integral that in this case $\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx.$

If $$a<b<c$$, we just showed that we can write $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx.\tag{i }$ Because we defined $\int_{{\color{red}b}}^{{\color{blue}c}}f(x)dx=-\int_{{\color{blue}c}}^{{\color{red}b}}f(x)dx,$ we can rewrite (i) as $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx-\int_{c}^{b}f(x)dx.$ Taking the last integral to the left, we get $\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\int_{a}^{b}f(x)dx.$ The truth of other cases (for example, if $$c<a<b$$) can be shown similarly.

Example 1
In an example in the previous section, we found $$\int_{1}^{3}x^{2}dx=\frac{26}{3}$$. Use the properties of the definite integral to evaluate $\int_{1}^{3}(2-5x^{2})dx.$

Solution 1

\begin{aligned} \int_{1}^{3}(2-5x^{2})dx &=\int_{1}^{3}2dx+\int_{1}^{3}-5x^{2}dx \qquad\small\text{(property (1))}\\ &=\int_{1}^{3}2dx-5\int_{1}^{3}x^{2}dx \qquad\small\text{(property (2))}\\ &=2(3-1)-5\times\frac{26}{3} \qquad\small\text{(property (3) and the given integral)}\\ & =-\frac{118}{3}.\end{aligned}

Example 2

Find $$\int_{8}^{9}f(x)dx$$, if we know that $$\int_{-1}^{9}f(x)dx=-2$$ and $$\int_{-1}^{8}f(x)dx=4$$.

Solution 2

Using property 4, we have $\underbrace{\int_{-1}^{9}f(x)dx}_{-2}=\underbrace{\int_{-1}^{8}f(x)dx}_{4}+\int_{8}^{9}f(x)dx.$ Therefore, $\int_{8}^{9}f(x)dx=-6.$

1. If $$f(x)\geq0$$ for $$a<x<b$$ then $\int_{a}^{b}f(x)dx\geq0.$
2. If $$f(x)\geq g(x)$$ for $$a<x<b$$ then $\int_{a}^{b}f(x)dx\geq\int_{a}^{b}g(x)dx.$

Property (5) says if the graph of $$f$$ does not lie below the $$x$$-axis, the signed area of the region lying between the curve $$y=f(x)$$ and the $$x$$-axis is nonnegative (see Figure 4).
Rigorously, if $$f(x)\geq0$$ then each term in $$\sum_{k=1}^{n}f(x_{k}^{*})(x_{k}-x_{k-1})$$ (for $$x_{k}>x_{k-1})$$ and its limit as $$n\to\infty$$ is nonnegative. Figure 4. If the graph of $$f$$ does not lie below the $$x$$-axis on $$[a,b]$$, the net signed area of the region under its curve is nonnegative.

The truth of property (6), because $$f(x)-g(x)\geq0$$, follows from properties 5, 1, and 2.

1. If $$m\leq f(x)\leq M$$ for $$a<x<b$$ then $m(b-a)\leq\int_{a}^{b}f(x)dx\leq M(b-a).$

For this property, see Figure 5.

Example 3

Show that $\pi\sqrt{2}\leq\int_{0}^{\pi}\sqrt{2+\sin x}\ dx\leq\pi\sqrt{3}.$

Solution 3

Because $$0\leq\sin x\leq1$$ on the interval $$[0,\pi]$$, the maximum value of $$\sqrt{2+\sin x}$$ on $$[0,\pi]$$ is $$\sqrt{2+1}=\sqrt{3}$$ and the minimum is $$\sqrt{2+0}=\sqrt{2}$$. That is, $\sqrt{2}\leq\sqrt{2+\sin x}\leq\sqrt{3}$ It follows from property (7) that $\underbrace{\sqrt{2}(\pi-0)}_{\approx4.44288}\leq\int_{0}^{\pi}\sqrt{2+\sin x}\ dx\leq\underbrace{\sqrt{3}(\pi-0)}_{\approx5.4414}.$ The value of the integral is $\int_{0}^{\pi}\sqrt{2+\sin x}\ dx\approx5.0922.$ Example 4

Use Property (7) to estimate the following integral $\int_{0.5}^{2}3^{1/x}dx.$

Solution 3

Because $3^{1/2}\leq3^{1/x}\leq3^{1/0.5}=3^{2}\qquad(0.5\leq x\leq2)$ we have $\sqrt{3}(2-0.5)\leq\int_{1}^{2}3^{1/x}dx\leq9(2-0.5)$ $2.59808\leq\int_{1}^{2}3^{1/x}dx\leq13.5$ The actual value of this integral is $$\approx4.57596.$$ 