5.4 Integration Formulas and the Net Change Theorem

[STARTS AT 4:04]

INSTRUCTOR: Let's consider the total displacement. Same scenario, I want to graph this, sketch it. At time 0 we're at negative 5. As I just said, 5/3, we're 0. And at time 3 the velocity is 4. 

So the velocity we're trying to find, or the distance we really want is these two triangles. The total distance is going to be summing those two, rather than subtracting like we did with the net. 

OK, let's go ahead and work that out. The total distance is equal to integral from 0 to 5/3 of 3t minus 5dt, plus the integral from 5/3 to 3, 3t minus 5dt. Now, because we have to add these two as just the quantity, we need to take absolute value bars. Take the absolute value of each of those. Which will not change much of this. 

So the antiderivative is 3t squared over 2 minus 5t. Evaluated from 0 to 5 over 3. And we'll take the absolute value of whatever we get there, plus 3 t squared over 2 minus 5t. Evaluate at 5/3 and 3. And we'll take the absolute value there. Now, notice the second one is going to be positive based on our graph. The first one is going to be negative. 

Now, evaluating each of those points, we get the absolute value of negative 25/6 plus, let's see that's, 19/6. So adding those two together-- that's not right. 19/6 is not right I don't think. 

2 plus 25 over 6. That's 8/3, actually. OK, so our total is a positive 41 over 6 meters. So again, the way I kind of describe the net versus the total is if I walk to my right 2 feet but then I go to my left 5 feet, where I end up is negative 3. I end up at 3 feet to my left. 

However, the total I walked was 7 feet. I forgot what I said. But the point is there that the total that I walked is the sum of those two distances, the absolute values of those. Whereas the net is a combination. One of those might be subtraction. But that's what it gets us.

[ENDS AT 7:36]