5.7E: Net Change Exercises
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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^2m/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.