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Some integration problems

  • Thread starter armolinasf
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Homework Statement



I'm having a bit of difficulty with these two integration problems:

1. Suppose [tex]\int^{2}_{0}[/tex]f(t)dt=3 calculate the following:

a) [tex]\int^{.5}_{0}[/tex]f(2t)dt

b) [tex]\int^{1}_{0}[/tex]f(1-t)dt

c) [tex]\int^{1.5}_{1}[/tex]f(3-2t)dt

The second problem is this:

If we assume that wind resistance is proportional to velocity, then the downward velocity, v, of a body of mass m falling vertically is given by:

v=(mg/k)(1-e[tex]^{(-kt)/m}[/tex])

where g is the acceleration due to gravity and k is a constant. Find the height, h, above the surface of the earth as a dunction of time. Assume the body starts at height h[tex]_{0}[/tex]



The Attempt at a Solution



For 1, I know that there is some sort of substitution that I'm just not seeing.

For 2, I basically treated it like a differential equation where v=dh/dt and I get as an antiderivative: (mg/k)(t-(m/k)e[tex]^{(k/m)t}[/tex]

I'm not sure how to get the height equal to 0 so that I can incorporate the h[tex]_{0}[/tex]

Any help is appreciated
 

Answers and Replies

  • #2
dextercioby
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Point 1. In your hypothesis, make a change of variable: t=2p. What do you get ?

Point 2. You may want to revise your integration. One of the signs is wrong.
 
  • #3
HallsofIvy
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Homework Statement



I'm having a bit of difficulty with these two integration problems:

1. Suppose [tex]\int^{2}_{0}[/tex]f(t)dt=3 calculate the following:
Are you sure this is correct? With this information, you can't come to any conclusion about the following integrals. If it were
[tex]\int_0^1 f(t)dt= 3[/tex]
then they are easy.

a) [tex]\int^{.5}_{0}[/tex]f(2t)dt

b) [tex]\int^{1}_{0}[/tex]f(1-t)dt

c) [tex]\int^{1.5}_{1}[/tex]f(3-2t)dt

The second problem is this:

If we assume that wind resistance is proportional to velocity, then the downward velocity, v, of a body of mass m falling vertically is given by:

v=(mg/k)(1-e[tex]^{(-kt)/m}[/tex])

where g is the acceleration due to gravity and k is a constant. Find the height, h, above the surface of the earth as a dunction of time. Assume the body starts at height h[tex]_{0}[/tex]



The Attempt at a Solution



For 1, I know that there is some sort of substitution that I'm just not seeing.

For 2, I basically treated it like a differential equation where v=dh/dt and I get as an antiderivative: (mg/k)(t-(m/k)e[tex]^{(k/m)t}[/tex]

I'm not sure how to get the height equal to 0 so that I can incorporate the h[tex]_{0}[/tex]

Any help is appreciated
?? Nothing is said about the height being 0. The height when t= 0 is [itex]h_0[/itex].
(Actually, two signs are wrong in your integral.)
 

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