1. ## Integration Confusion

Doing this problem that involves distance a satellite travels with respect to earth (in thousands of miles). Find the total distance traveled from 0 to 5 hours.

$s(t) = t^3 - 2t^2 - 4T + 12 (position)$

$s'(t) = 3t^2 -4t -4$

Critical Points: [-2/3, 2]

Now I am trying to gain a more tacit understanding of calculus, so my question is:
why is the integration of the velocity (position) from 0 to 5 hours not equal to the sum of integrals between [0,2] and [2,5]? I thought that integration is summation, so why is it then when an object is traveling away from earth, you have to do a separate integral for it?

The answer is 71,000 miles, but I can't seem to comprehend why integral doesn't count the "negative" velocity in total distance, but the area summation does.

Sorry if this question is confusing, appreciate any help .

-Warren

2. Originally Posted by Warrenx
Doing this problem that involves distance a satellite travels with respect to earth (in thousands of miles). Find the total distance traveled from 0 to 5 hours.

$s(t) = t^3 - 2t^2 - 4T + 12 (position)$

$s'(t) = 3t^2 -4t -4$

Critical Points: [-2/3, 2]

Now I am trying to gain a more tacit understanding of calculus, so my question is:
why is the integration of the velocity (position) from 0 to 5 hours not equal to the sum of integrals between [0,2] and [2,5]? I thought that integration is summation, so why is it then when an object is traveling away from earth, you have to do a separate integral for it?

The answer is 71,000 miles, but I can't seem to comprehend why integral doesn't count the "negative" velocity in total distance, but the area summation does.

Sorry if this question is confusing, appreciate any help .

-Warren
Could you post the exact question

CB

3. I am reading "Calculus for Idiots" and I currently on chapter 16: Applications of the Fundamental Theorem. This is the problem:

Problem 3: When satellites circle closely around a planet or moon, the gravitational field surrounding the celestial body both increases the satellite's velocity and changes its direction in an orbital move called a "slingshot." (As you may know from the movie Apollo 13, Tom Hanks and his crew executed a slingshot maneuver around the moon to hurl themselves back toward earth.) Let's say that a ship is executing this maneuver has position equation s(t) = t^3 -2t^2 -4t +12, where t is in hours, and s(t) represents thousands of miles from earth. What is the total distance traveled by the craft during the first five hours?

I just want to understand why it is that if the velocity of the s(t) is negative, why do you have to make that a negative integral, if its summation isn't it cumulative, so it would add both the distance to and from earth in fell swoop?

Thanks

-Warren

4. Originally Posted by Warrenx
I am reading "Calculus for Idiots" and I currently on chapter 16: Applications of the Fundamental Theorem. This is the problem:

Problem 3: When satellites circle closely around a planet or moon, the gravitational field surrounding the celestial body both increases the satellite's velocity and changes its direction in an orbital move called a "slingshot." (As you may know from the movie Apollo 13, Tom Hanks and his crew executed a slingshot maneuver around the moon to hurl themselves back toward earth.) Let's say that a ship is executing this maneuver has position equation s(t) = t^3 -2t^2 -4t +12, where t is in hours, and s(t) represents thousands of miles from earth. What is the total distance traveled by the craft during the first five hours?

I just want to understand why it is that if the velocity of the s(t) is negative, why do you have to make that a negative integral, if its summation isn't it cumulative, so it would add both the distance to and from earth in fell swoop?

Thanks

-Warren
If I travel north for 1 hour at 12 mph and then travel south for 1 hour at 12 mph, how far have I gone?

CB

5. Originally Posted by Warrenx
Now I am trying to gain a more tacit understanding of calculus, so my question is:
why is the integration of the velocity (position) from 0 to 5 hours not equal to the sum of integrals between [0,2] and [2,5]?
When you integrate a function that is below the x-axis, you will get a negative answer. If you integrate a function above the x-axis, you will get a positive answer.
By integrating from 0 to 5, you mix up the positive and negative, thus the negative will cancel out the positive and you will get a smaller answer.

Originally Posted by Warrenx
I thought that integration is summation, so why is it then when an object is traveling away from earth, you have to do a separate integral for it?

The answer is 71,000 miles, but I can't seem to comprehend why integral doesn't count the "negative" velocity in total distance, but the area summation does.
When you integrate a velocity function, you get displacement, not distance. Velocity is a vector, so when you integrate velocity, you get a vector called displacement. The graph of this function shows the displacement of the object. Distance is scalar, therefore it does not account for positive or negative, but displacement is a vector, thus it accounts for the negative/positive position.

When you integrate it you get a displacement function so to stop the negative from cancelling out the positive, you have to seperate out the integral.