5.7E: Net Change Exercises

5.7: Net Change Exercises

Use basic integration formulas to compute the following antiderivatives.

207) $\displaystyle ∫(\sqrt{x}−\frac{1}{\sqrt{x}})dx$

Answer:

$\displaystyle ∫(\sqrt{x}−\frac{1}{\sqrt{x}})dx=∫x^{1/2}dx−∫x^{−1/2}dx=\frac{2}{3}x^{3/2}+C_1−2x^{1/2+}C_2=\frac{2}{3}x^{3/2}−2x^{1/2}+C$

208) $\displaystyle ∫(e^{2x}−\frac{1}{2}e^{x/2})dx$

209) $\displaystyle ∫\frac{dx}{2x}$

Answer:

$\displaystyle ∫\frac{dx}{2x}=\frac{1}{2}ln|x|+C$

210) $\displaystyle ∫\frac{x−1}{x^2}dx$

211) ​​​​​​​ $\displaystyle ∫^π_0(sinx−cosx)dx$

Answer:

$\displaystyle ∫^π_0sinxdx−∫^π_0cosxdx=−cosx|^π_0−(sinx)|^π_0=(−(−1)+1)−(0−0)=2$

212) $\displaystyle ∫^{π/2}_0(x−sinx)dx$

NET CHANGE

223) Suppose that a particle moves along a straight line with velocity $\displaystyle v(t)=4−2t,$ where $\displaystyle 0≤t≤2$ (in meters per second). Find the displacement at time t and the total distance traveled up to $\displaystyle t=2.$

Answer:

$\displaystyle d(t)=∫^t_0v(s)ds=4t−t^2$. The total distance is $\displaystyle d(2)=4m.$

224) Suppose that a particle moves along a straight line with velocity defined by $\displaystyle v(t)=t^2−3t−18,$ where $\displaystyle 0≤t≤6$ (in meters per second). Find the displacement at time t and the total distance traveled up to $\displaystyle t=6.$

225) Suppose that a particle moves along a straight line with velocity defined by $\displaystyle v(t)=|2t−6|,$ where $\displaystyle 0≤t≤6$ (in meters per second). Find the displacement at time t and the total distance traveled up to $\displaystyle t=6.$

Answer:

$\displaystyle d(t)=∫^t_0v(s)ds.$ For $\displaystyle t<3,d(t)=∫^t_0(6−2t)dt=6t−t^2$. For $\displaystyle t>3,d(t)=d(3)+∫^t_3(2t−6)dt=9+(t^2−6t)$. The total distance is $\displaystyle d(6)=9m.$

226) Suppose that a particle moves along a straight line with acceleration defined by $\displaystyle a(t)=t−3,$ where $\displaystyle 0≤t≤6$ (in meters per second). Find the velocity and displacement at time t and the total distance traveled up to $\displaystyle t=6$ if $\displaystyle v(0)=3$ and $\displaystyle d(0)=0.$

227) A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the velocity $\displaystyle v(t)$ and the height $\displaystyle h(t)$ of the ball t seconds after it is thrown and before it returns to the ground.

Answer:

$\displaystyle v(t)=40−9.8t;h(t)=1.5+40t−4.9t^2$m/s

228) A ball is thrown upward from a height of 3 m at an initial speed of 60 m/sec. Acceleration resulting from gravity is $\displaystyle −9.8 m/sec^2$. Neglecting air resistance, solve for the velocity $\displaystyle v(t)$ and the height $\displaystyle h(t)$ of the ball t seconds after it is thrown and before it returns to the ground.