First Application of Definite Integral

Practice 1: Using geometry (Fig. 17): $\mathrm{A}=\frac{1}{2}(2)(1)=1 \text { and } \mathrm{B}=\frac{1}{2}(4)(2)=4$ so $total area = A+B=5$.

Using integrals:

$A=\int_{0}^{1}(3-x)-(1+x) d x=\int_{0}^{1}(2-2 x) d x=2 x-\left.x^{2}\right|_{0} ^{1}=(2-1)-(0)=1$

$B=\int_{1}^{3}(1+x)-(3-x) \mathrm{dx}=\int_{1}^{3}(2 x-2) d x=x^{2}-\left.2 x\right|_{1} ^{3}=(9-6)-(1-2)=4$

The single integral $\int_{0}^{3}(1+x)-(3-x) d x$ is not correct: $\int_{0}^{3}(1+x)-(3-x) d x=3$.

Practice 2: $Average value = \frac{1}{\mathrm{~b}-\mathrm{a}} \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}=\frac{1}{9-0} \int_{0}^{9} 5+\sqrt{t} \mathrm{dt}$

$=\frac{1}{9} \int_{0}^{9} 5+\mathrm{t}^{1 / 2} \mathrm{dt}=\left.\frac{1}{9}\left(5 \mathrm{t}+\frac{2}{3} \mathrm{t}^{3 / 2}\right)\right|_{0} ^{9}$

$=\frac{1}{9}\left(45+\frac{2}{3} 9^{3 / 2}\right)-\frac{1}{9}(0)=\frac{1}{9}(45+18)=7$ cars per hour.

Practice 3: $\text { Work }=(\text { force }) \cdot(\text { distance })=(10 \text { pounds }) \cdot(30 \text { feet })=\mathbf{3 0 0}  foot-pounds$.

Practice 4: (a) force = (force for cable) + (force for object)

= (length of cable)(density of cable) + 10 pounds

$=(50-x \text { feet })(3 \text { pounds } / \text { foot })+10 \text { pounds }=160-3 \mathbf{x}$ pounds.

(b) "from the ground to a height of 10 feet:"

$\text { Work } \approx \sum(\text { subinterval work })=\sum \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \Delta \mathrm{x}_{\mathrm{i}}=\sum\left\{160-3 \mathrm{c}_{\mathrm{i}}\right\} \Delta \mathrm{x}_{\mathrm{i}}$ (a Riemann sum)

$\longrightarrow \int_{0}^{10}\{160-3 x\} d x$ as the mesh approaches 0.

"From a height of 10 feet to a height of 20 feet:" work = $\int_{10}^{20}\{160-3 x\} d x$.