
5.7E: Net Change Exercises


5.7: Net Change Exercises

Use basic integration formulas to compute the following antiderivatives.

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

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

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

209) $$\displaystyle ∫\frac{dx}{2x}$$

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

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

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

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

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 $$\displaystyle t=2.$$

$$\displaystyle d(t)=∫^t_0v(s)ds=4t−t^2$$. 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 $$\displaystyle t=6.$$

$$\displaystyle d(t)=∫^t_0v(s)ds.$$ 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.
$$\displaystyle v(t)=40−9.8t;h(t)=1.5+40t−4.9t^2$$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.