Practice 1: Using geometry (Fig. 17):
so
.
Using integrals:
![A=\int_{0}^{1}(3-x)-(1+x) d x=\int_{0}^{1}(2-2 x) d x=2 x-\left.x^{2}\right|_{0} ^{1}=(2-1)-(0)=1 A=\int_{0}^{1}(3-x)-(1+x) d x=\int_{0}^{1}(2-2 x) d x=2 x-\left.x^{2}\right|_{0} ^{1}=(2-1)-(0)=1](../../../../filter/tex/pix.php/f1ab945ea6d4d18edb9eb3eb9ca6f15d.svg)
![B=\int_{1}^{3}(1+x)-(3-x) \mathrm{dx}=\int_{1}^{3}(2 x-2) d x=x^{2}-\left.2 x\right|_{1} ^{3}=(9-6)-(1-2)=4 B=\int_{1}^{3}(1+x)-(3-x) \mathrm{dx}=\int_{1}^{3}(2 x-2) d x=x^{2}-\left.2 x\right|_{1} ^{3}=(9-6)-(1-2)=4](../../../../filter/tex/pix.php/4740fe99969dfc0429c7faaaa2c8f40c.svg)
The single integral
is not correct:
.
![](../../../../pluginfile.php/3640352/mod_book/chapter/21793/Fig%2017.png)
Practice 2: ![Average value = \frac{1}{\mathrm{~b}-\mathrm{a}} \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}=\frac{1}{9-0} \int_{0}^{9} 5+\sqrt{t} \mathrm{dt} Average value = \frac{1}{\mathrm{~b}-\mathrm{a}} \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}=\frac{1}{9-0} \int_{0}^{9} 5+\sqrt{t} \mathrm{dt}](../../../../filter/tex/pix.php/7d88d2be53be56579575d5eb2d4fd68e.svg)
![=\frac{1}{9} \int_{0}^{9} 5+\mathrm{t}^{1 / 2} \mathrm{dt}=\left.\frac{1}{9}\left(5 \mathrm{t}+\frac{2}{3} \mathrm{t}^{3 / 2}\right)\right|_{0} ^{9} =\frac{1}{9} \int_{0}^{9} 5+\mathrm{t}^{1 / 2} \mathrm{dt}=\left.\frac{1}{9}\left(5 \mathrm{t}+\frac{2}{3} \mathrm{t}^{3 / 2}\right)\right|_{0} ^{9}](../../../../filter/tex/pix.php/48934150fe6286fcdcdfaba306e47636.svg)
cars per hour.
Practice 3:
.
Practice 4: (a) force = (force for cable) + (force for object)
= (length of cable)(density of cable) + 10 pounds
pounds.
(b) "from the ground to a height of 10 feet:"
(a Riemann sum)
as the mesh approaches 0.
"From a height of 10 feet to a height of 20 feet:" work =
.