 Let's look at an example of a motion problem that's going to require us to use integrals and let's try to calculate the displacement or the net change of position associated to a particle that's moving with the following velocity function. So its velocity of v of t is given as t squared minus t minus six. We're measuring this in meters per second. So by the by the theorem of calculus we actually know that since velocity is the derivative of position if we want to find out the change of the change of position this this displacement if we want to figure out the displacement of the particle during the time period one to four this displacement s of four minus s of one this will equal the integral from one to four of the velocity function. Now we don't have enough information to know what the position function is because we would need some information about where is it located at time equals zero or something like that. We need some initial value but with just the velocity at hand we actually can calculate the net change of position the displacement of this thing. So s of four minus s of one this we can figure out how far it moved in this time period this is coming to integral from one to four of the velocity function and so using the velocity function they have up there we get t squared minus t minus six dt by properties of integration and by the power rule for anti-derivatives we'll get t cubed over three minus t squared over two minus six t we evaluate from one to four and now we hit the most arduous part of the whole calculation here the arithmetic we plug in four so we're gonna get four cubed over three minus four squared over two minus six times four and then we have to subtract from that when we plug in one so we're gonna get one cubed over three minus one squared over two minus six times one simplifying these things six four cubed is a 64 four squared is an eight or is a 16 which we get 16 halves right there six times four is 24 so that's our first group for the second one we get one third one half and six combining like terms we could take away a third from 64 thirds which gives us 63 thirds we can add a half to negative 16 halves which gives us a negative 15 halves and then lastly we're gonna take negative 24 plus six which is a negative 18 right there and then continuing to combine some like terms we get some fractions here uh 60 well i guess 63 uh 63 thirds we could we could simplify that thing uh a little bit more couldn't we uh three goes into 63 21 times like so 21 take away 18 would give us three three minus 15 halves it's somewhat unavoidable in that situation we get six over halves take away 15 halves and then that gives us a difference of negative nine halves like so and this represents the displacement of our particle and here we get a negative value we actually do want to keep it negative i know in previous videos we saw that when you're looking for the geometric area we want that to be positive in this situation we want this to be a negative value and so this would tell us that the displacement is equal to negative four and a half meters the negative is important because it gives us a direction it tells us that it is uh since it's a negative 4.5 it's to the left of the quote unquote zero location and so calculating displacement just comes down to computing an integral there's not much more to it than that um suppose on the other hand though we want to find the distance traveled by the particle during this time period we want to find the total distance in that situation the total distance is actually going to be the integral from one to four of the absolute value of the whole velocity function dt like so and so how does absolute value affect how we would compute this well we would want to be looking for an anti derivative of our function when we take the absolute value and putting the absolute value in there kind of complicates the anti-derivative process but there's a nice little trick one could do to sneak around that if we were to graph our velocity function so you graph it with respect to time right here notice velocity uh v of t we had before was t squared minus t minus six if we were to factor that thing uh this thing factors uh will factors of negative six and out of t negative one we can take t minus three and t plus two and so we'll get we'll get two uh these would actually represent the critical numbers of the original position function here we're going to get three and negative two so if we're only interested in the domain one to four you'll notice that the number three falls inside that region okay and so what we're going to do is we're going to break up our integral into pieces based upon this number three right here so we're going to integrate from one to three absolute value of v of t dt and then we add to that the integral from three to four of the absolute value of v of t dt and so why we break it up that way is that our velocity function it's going to look something like the following uh if we come over here it was it was somewhere it was somewhere above the x axis and it crosses at three uh to somewhere below the x axis right here i guess i actually take it back to probably the other way around our function's going to look something like this it crosses the x axis at three like so and so the area that's under the curve when you're to the left of three is going to be positive but the area to you went to the right of three excuse me but when you're to the left of three it's going to be negative area right here and so we can break it up because there's no x intercepts between the between one and three and three and four if we take absolute value it's going to be all negative or all positive so the absolute value let's turn everything to be positive in the end and so then we can actually take the absolute value outside of the integral process we integrate from one to three vt dt and then we also integrate from three to four vt dt and this is the same velocity function we have before this t squared minus t minus six so like we saw on the other saw on the other slide here if we integrate from one to three we're going to get t cubed over three minus t squared over two minus six t as we go from one to three absolute value and then we also do this for the absolute value then the x one integrates the exact same anti-derivative t cubed over three minus t squared over two minus six t as you go from this time three to four like so so you have to break it up into two pieces because this first region right here we're actually going to get that this was going backward because the velocity was negative right there and then the other one this is when it was moving forward in the first calculation we did displacement the forward part cancels with some of the backward part so we only got the net change which is a negative 4.5 meters this time we want to consider the whole the whole situation our particle was moving backwards and then starts to move forward a little bit and so the previous calculation was only looking at the net change in position right there but this time we're interested in figuring out what is the length of this entire journey like so and so to do this calculation we got to plug in we got to plug in the the in the one the three so we have this absolute value we got three cubed over three minus three squared over two minus six times three minus from that one third minus a half minus six right there that's in the first absolute value and then in the second absolute value we're going to have 64 over three minus 16 over two minus 24 minus this three cubed over three minus three squared over two minus the six times three you haven't computed that one yet it's all since inside the absolute value I do want to point out though that this expression right here is identical to this one but they don't cancel out because they're separated by these absolute values we can't just throw them together now I'm not going to bore us with the details of the arithmetic this time around but if you do the first set of absolute values you'll end up with a negative 44 over six and with the second one you end up with the absolute value of 17 over six and so like I said this first distance was moving backwards which is why we got a negative this one was moving forward which is why it's positive taking the absolute value we get 44 over six plus 17 over six or in other words 61 over six or approximately 10.17 meters was the total distance traveled by this particle which is a lot bigger than the absolute value we saw last time 4.5 and that's because a lot of the lot of the journey of this particle actually doubled back on where it already was and so the displacement's not going to see that the displacement is just the regular definite integral and it gives you this net change if we want to know the total distance what we have to do is take the integral of the absolute value of the function and because of the absolute value you want to break it up the function v of t along its x intercepts and treat each piece differently making sure that you take the absolute values each of those pieces focusing of course where are the x intercepts we took our interval remember one to four and we had to break it up at three that is taking the integral from one to three and then from three to four so net change pretty simple just calculate an integral for total change uh gross change in that case you have to find the x intercepts and then integrate each of the intervals that are divided into smaller pieces using these x intercepts