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.
