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Use the properties of the definite integral to express the definite integral of $f\left(x\right)=\mathrm{-3}{x}^{3}+2x+2$ over the interval $\left[\mathrm{-2},1\right]$ as the sum of three definite integrals.
Using integral notation, we have ${\int}_{\mathrm{-2}}^{1}\left(\mathrm{-3}{x}^{3}+2x+2\right)dx}.$ We apply properties 3. and 5. to get
Use the properties of the definite integral to express the definite integral of $f\left(x\right)=6{x}^{3}-4{x}^{2}+2x-3$ over the interval $\left[1,3\right]$ as the sum of four definite integrals.
$6{\displaystyle {\int}_{1}^{3}{x}^{3}dx}-4{\displaystyle {\int}_{1}^{3}{x}^{2}dx}+2{\displaystyle {\int}_{1}^{3}xdx}-{{\displaystyle \int}}_{1}^{3}3dx$
If it is known that ${\int}_{0}^{8}f\left(x\right)dx}=10$ and ${\int}_{0}^{5}f\left(x\right)dx}=5,$ find the value of ${\int}_{5}^{8}f\left(x\right)dx}.$
By property 6.,
Thus,
If it is known that ${\int}_{1}^{5}f\left(x\right)dx}=\mathrm{-3$ and ${\int}_{2}^{5}f\left(x\right)dx}=4,$ find the value of ${\int}_{1}^{2}f\left(x\right)dx}.$
−7
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say that if a function $f\left(x\right)$ is above another function $g\left(x\right),$ then the area between $f\left(x\right)$ and the x -axis is greater than the area between $g\left(x\right)$ and the x -axis. This is true depending on the interval over which the comparison is made. The properties of definite integrals are valid whether $a<b,a=b,$ or $a>b.$ The following properties, however, concern only the case $a\le b,$ and are used when we want to compare the sizes of integrals.
Compare $f\left(x\right)=\sqrt{1+{x}^{2}}$ and $g\left(x\right)=\sqrt{1+x}$ over the interval $\left[0,1\right].$
Graphing these functions is necessary to understand how they compare over the interval $\left[0,1\right].$ Initially, when graphed on a graphing calculator, $f\left(x\right)$ appears to be above $g\left(x\right)$ everywhere. However, on the interval $\left[0,1\right],$ the graphs appear to be on top of each other. We need to zoom in to see that, on the interval $\left[0,1\right],g\left(x\right)$ is above $f\left(x\right).$ The two functions intersect at $x=0$ and $x=1$ ( [link] ).
We can see from the graph that over the interval $\left[0,1\right],g\left(x\right)\ge f\left(x\right).$ Comparing the integrals over the specified interval $\left[0,1\right],$ we also see that ${\int}_{0}^{1}g\left(x\right)dx}\ge {\displaystyle {\int}_{0}^{1}f\left(x\right)dx$ ( [link] ). The thin, red-shaded area shows just how much difference there is between these two integrals over the interval $\left[0,1\right].$
We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus,
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