# Integrating: mg/b(1 - e^(-bt/m)

• jmwachtel
In summary, Homework Statement: The student is trying to integrate mg/b(-b/m * e^(-bt/m)) but is having trouble with the exponential part of the equation. They get help from a classmate who explains that they need to use the chain rule for integration to solve for mgt/b. However, even with the help of a classmate, the student still does not get the integral correct. It seems that something is wrong with their setup for the problem, and they need more help to figure out what is wrong.
jmwachtel

## Homework Statement

I need help integrating: mg/b(1 - e^(-bt/m)

## The Attempt at a Solution

I came up with mgt/b - mg/b(-b/m * e^(-bt/m)

It's been a while since calculus, can someone help me with this.

If this is the integral you mean:

$$\frac{mg}{b}\int(1-e^{-bt/m})dt$$

then you are correct in your integration.

That is it, but the numbers are coming out negative, they should be positive. Something is not right. I know I have to take it from 0 to t, but I'm not sure what to do with it. Please help.

Anyone else know calculus?

Can you give more information on the problem? It could simply be that you've take a particular direction to be positive, and the answer in the back of the book assumes the opposite direction is positive.

I get the integral to be

= mg/b * {[t-(m/b)exp(-bt/m)]}
= {mgt/b + [(m^2)g/(b^2)]exp(-bt/m)}
= {mgt/b + mg/b(m/b * e^(-bt/m)}

you got the exponential integration wrong:
Integral of: exp(kx)dx = {1/(dkx/dx)exp(kx)} = {(1/k)exp(kx)}
in this case k = -b/m
so integral of exp[(-b/m)t] = (-m/b).exp[(-b/m)t]

Last edited:
The answer should be around .3 positive when T is equal to .05. I get a negative -.02417. The integral is still not right even with the one above, what is going on?

and: m = .001, g=-9.81, b=.025

The integral you posted first is not correct as cynapse explained, even in corrected form it's an indefinite integral with the constant of integration missing. It sounds like you need a definite integral.
Did you get the right definite integral for the limits of integration 0 and T?
The answer for the equation as you described it is -0.0084.
This leads me to believe that you don't have the correct set up for the problem.
Are the units consistent?
Do you have the correct limits of integration?
If you post the full wording of the problem as well as units of known quantity's with a more complete description of what you are doing I can try to help you more.

Last edited:
Everything is correct, we are setting up a spreadsheet in class. We use this for Analytical
approach. v is calculated using the equation above, so to get back to y, you have to integrate. I know what y should be which I stated above because that's through the numerical approach.

It seems you're finding distance traveled while experiencing air resistance. Just do this:
$$\frac{mg}{b}\int(1-e^{-bt/m})dt$$
Then, integrate each term in the integral separately in respect to time t (find the integral of 1 and the integral of -e^(-bt/m) individually). For the integral of -e^(-bt/m), use (-bt/m) as a function u. Then simply use the chain rule for integration to solve the problem.

Here is what I have got and it's still not working, this is driving me nuts!

=(((m*g/b)*t)+((m*g/b)*((m/b)*EXP(-b*t/m))))-(m^2*g/b^2)

What am I doing wrong?

That's the correct integral for the limits 0 & T.
At least I think it is. I would recommend using the forums handy mathematical formatting so the equations are easier to read.
So I still think something else is wrong here.

This still not solved. I can't figure out what I am doing wrong. Where is the equation formatting?

[$$\frac{mgt}{b}$$ + $$\frac{mg}{b}$$ ($$\frac{m}{b}$$ e$$^{\frac{-bt}{m}}$$)] - [$$\frac{m^{2}g}{b^{2}}$$]

This is integration I have come up with for the equation listed above that is not working! IS my integration wrong? This from 0 to t.

That is the right integral for the equation G01 posted for the lower limit of 0 and an upper limit of t.
Side note: your equations will look nicer and be easier to work with if you simplify them.

$$\frac{mg}{b}$$$$[T + \frac{m}{b}(e^\frac{-bT}{m} - 1)]$$

Ok here is my teachers reply:

Your integration is still incorrect. See attachment for explanation. You MUST have correctly corresponding limits on the integrals. At time=0, y=yo; at time=t, y=y.

I don't understand. The attached says:

v = dy/dt

dy = vdt

$$\int$$dy (from yintial to y) = $$\int$$vdt (from 0 to t)

Anyone understand this?

## 1. What does the equation "Integrating: mg/b(1 - e^(-bt/m)" represent?

The equation represents the integration of a function that describes the position of an object subjected to a constant force and drag force.

## 2. How is the equation derived?

The equation is derived by solving the differential equation for the motion of an object under constant force and drag. This involves using the fundamental theorem of calculus and applying the appropriate integration techniques.

## 3. What is the significance of the variables in the equation?

The variable m represents the mass of the object, g is the gravitational constant, b is the drag coefficient, t is the time, and the variable e is the base of the natural logarithm. The equation also includes the constant b/m, which is related to the time constant of the system.

## 4. How is the equation used in scientific research?

The equation is commonly used in physics and engineering to model the motion of objects under the influence of gravity and drag. It can also be used to analyze data and make predictions about the behavior of physical systems.

## 5. Are there any limitations to the use of this equation?

Yes, the equation assumes a constant force and drag throughout the motion of the object. It also does not take into account other external forces or factors such as air resistance or friction. Additionally, it may not accurately model more complex systems with changing forces or drag coefficients.

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