How Do You Solve These Challenging Integration Problems?

[armolinasf]

Homework Statement

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

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

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

b) \int^{1}_{0}f(1-t)dt

c) \int^{1.5}_{1}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^{(-kt)/m})

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_{0}

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^{(k/m)t}

I'm not sure how to get the height equal to 0 so that I can incorporate the h_{0}

Any help is appreciated

 

[dextercioby]

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.

 

[HallsofIvy]

Homework Statement

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

1. Suppose \int^{2}_{0}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
\int_0^1 f(t)dt= 3
then they are easy.

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

b) \int^{1}_{0}f(1-t)dt

c) \int^{1.5}_{1}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^{(-kt)/m})

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_{0}

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^{(k/m)t}

I'm not sure how to get the height equal to 0 so that I can incorporate the h_{0}

Any help is appreciated

?? Nothing is said about the height being 0. The height when t= 0 is h_0.
(Actually, two signs are wrong in your integral.)