Practice Problems

1.

(a) See Figure.

(b) $\mathrm{A}(1)=0, \mathrm{~A}(2)=1.5, \mathrm{~A}(3)=4, \mathrm{~A}(4)=6.5$

(c) $\text { (c) } \mathrm{A}^{\prime}(1)=1, \mathrm{~A}^{\prime}(2)=2, \mathrm{~A}^{\prime}(3)=3, \mathrm{~A}^{\prime}(4)=2$

3.

(a) see Figure

(b) $\mathrm{A}(1)=0, \mathrm{~A}(2)=0.5, \mathrm{~A}(3)=0, \mathrm{~A}(4)=-1$

(c) $\mathrm{A}^{\prime}(1)=1, \mathrm{~A}^{\prime}(2)=0, \mathrm{~A}^{\prime}(3)=-1, \text { A }^{\prime}(4)=-1$

5.

(a) see Figure

(b)$\mathrm{A}(1)=0, \mathrm{~A}(2)=2, \mathrm{~A}(3)=4, \mathrm{~A}(4)=6$

(c) $A^{\prime}(1)=2, A^{\prime}(2)=2, A^{\prime}(3)=2, A^{\prime}(4)=2$

7.

(a) see Figure

(b) $\mathrm{A}(1)=0, \mathrm{~A}(2)=4.5, \mathrm{~A}(3)=8, \mathrm{~A}(4)=10.5$

(c) $A^{\prime}(1)=5, A^{\prime}(2)=4, A^{\prime}(3)=3, A^{\prime}(4)=2$

9. $\left.\mathrm{x}^{2}\right|_{0} ^{3}=9,\left.\mathrm{x}^{2}\right|_{1} ^{3}=8,\left.\mathrm{x}^{2}\right|_{0} ^{1}=1$

11. $\left.2 \mathrm{x}^{3}\right|_{1} ^{3}=52,\left.2 \mathrm{x}^{3}\right|_{1} ^{2}=14,\left.2 \mathrm{x}^{3}\right|_{0} ^{3}=54$

13. $\left.\mathrm{x}^{4}\right|_{0} ^{3}=81,\left.\mathrm{x}^{4}\right|_{1} ^{3}=80,\left.\mathrm{x}^{4}\right|_{0} ^{1}=1$

15. $\left.\mathrm{x}^{3}\right|_{-3} ^{3}=54,\left.\mathrm{x}^{3}\right|_{-3} ^{0}=27,\left.\mathrm{x}^{3}\right|_{0} ^{3}=27$

17. $\left.x^{3}\right|_{0} ^{2}=8,\left.x^{3}\right|_{1} ^{3}=26$

19.

(a) distance = $\int_{0}^{10} 2 \mathrm{t} \mathrm{dt}=\left.\mathrm{t}^{2}\right|_{0} ^{10}=100$ feet.

(b) Find $T$ so $50=\int_{0}^{\mathrm{T}} 2 \mathrm{t} \mathrm{dt}=\left.\mathrm{t}^{2}\right|_{0} ^{\mathrm{T}}=\mathrm{T}^{2} . \mathrm{T}=\sqrt{50} \approx 7.07$ seconds.

21.

(a) distance = $\int_{0}^{10} 4 \mathrm{t}^{3} \mathrm{dt}=\left.\mathrm{t}^{4}\right|_{0} ^{10}=10,000$ feet.

(b) $5000=\int_{0}^{T} 4 t^{3} \mathrm{dt}=\mathrm{T}^{4} \cdot \mathrm{T}=\sqrt[4]{5000} \approx 8.41 \text { seconds }$.

23.

(a) velocity = $75-3 t^{2}=0$ when $t = 5$ seconds. (b) distance = $\int_{0}^{5} 75-3 t^{2} \mathrm{dt}=75 \mathrm{t}-\left.\mathrm{t}^{3}\right|_{0} ^{5}=250$ feet.

(b) $125=\int_{0}^{T} 75-3 \mathrm{t}^{2} \mathrm{dt}=75 \mathrm{t}-\left.\mathrm{t}^{3}\right|_{0} ^{\mathrm{T}}=75 \mathrm{~T}-\mathrm{T}^{3} \text { so } \mathrm{T}^{3}-75 \mathrm{~T}+125=0$ (solve using Newton's method or by examining the graph of $y=x^{3}-75 x+125$) and $\mathrm{T} \approx 1.74$ seconds.

25. The total area is $\int_{0}^{3} x^{2} d x=\left.\frac{1}{3} x^{3}\right|_{0} ^{3}=9$.

(a) Find $T$ so $\frac{1}{2} \cdot 9=\frac{9}{2}=\int_{0}^{\mathrm{T}} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{27 / 2} \approx 2.38$.

(b) Find $T$ so $\frac{1}{3} \cdot 9=3=\int_{0}^{T} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{9} \approx 2.08$.

Then find $T$ so $\frac{2}{3} \cdot 9=6=\int_{0}^{\mathrm{T}} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{18} \approx 2.62$.