Calculus - The Fundamental Theorem, Part 1

now I'm going to talk about the fundamental theorem of calculus and I'll start by stating the theorem and then we'll work a couple of simple examples and then I'll come back after that and explain to you why the theorem has to be true and I'll be giving a geometrical argument that will hopefully elucidate some of the important points about it so first of all the statement of the theorem the fundamental theorem of calculus states that the integral from A to B of f of X DX equals and before I say what it equals let me say this when you see that the integral from A to B of f of X DX what should automatically pop into your head is a mental picture something like this some x set of XY axes with some function f of X and then two points a and B and that's these points here a and B and those points serve as a left and right boundary or a lower limit in an upper limit for this area so the area bound by the graph and the axis and in those a and B values that area is this the integral from A to B of f of X DX and the fundamental theorem of calculus says that that area is equal to G of B minus G of a and this function G that just showed up G is the antiderivative of this function f so we'll take note of that it's equal to G of B minus G of a where G is the antiderivative is the antiderivative of F so this this theorem isn't just a mathematical statement it tells us how to calculate this area if we want to find the area underneath this curve right here between these two points what we need to do is take this function f and find the antiderivative the antiderivative of that function will be another function and we've called it G and then we just find function G evaluated at point B and function G evaluated at point a and we subtract and that will give us this exact area okay let's look at an example here we're told find the area under the graph of f of x equals x squared from x equals 1 to x equals 2 so this is a simple graph this parabola so let's let's draw it I'll just put some points on the graph here X 1 2 3 4 and my y-axis I'll mark it off 1 2 3 & 4 and then let's plot some points this parabola we know goes to 0 0 1 1 & 2 4 pretty simple well known graph and it looks something like this ok that's the parabola f of x equals x squared and we're looking for this area from 1 to 2 okay the area under the graph that area and the top edge of this shape here is not a straight line that's a curve right there so we can't just calculate the area of a trapezoid because it's not a perfect trapezoid but here's what we're going to do we're going to find the antiderivative of this function so what's the antiderivative of x squared that be the integral of x squared DX well that's pretty easy it's X cubed over 3 and that's what we call G in our statement of the theorem so I evaluate function G at x equals 2 and at x equals 1 and then I subtract so I'm going to find G of 2 minus G of 1 and that's pretty easy G of 2 is just going to be 2 cubed over 3 and G of 1 is just going to be 1 cubed over 3 so it's 2 cubed over 3 minus 1 cubed over 3 that's 8 thirds minus 1/3 and that's our answer seven thirds and that was pretty easy to do and not only that but that's the exact area right there it's not an approximation that is exactly the correct answer and we found it and I'll show you a notation here that's important to know because it's very very common this evaluation of this function from from one to two the area from one to two it's commonly written like this I would write X cubed over three you put a vertical line and you write a 1 and a 2 and these are what we call the lower limit and the upper limit and all this means is exactly what we did this means put the put the variable 2 in for X and evaluate the function and and then put the value 1 in for X and evaluate the function so this is exactly what we just did 2 cubed over 3 minus 1 cubed over 3 and it came out very nicely to 7/3 and that's easy to do that's a lot easier than chopping it up into a lot of little pieces doing all these calculations for a Riemann sum or for a trapezoid and not only are those a lot of calculations but they're in exact this is a very simple calculation and it's exact and it's easy to do as long as we can find this function the antiderivative of the given curve here's one more example starting from rest a rocket accelerates for seven seconds and the acceleration is given by this function one point two T squared now note that this is non constant acceleration and we're asked how fast is the rocket moving at the end of the seven seconds now before I solve this and this is going to be pretty quick to solve let me say something about the physics here you might remember some equations from your physics class that look something like this there's an equation for position X is equal to X zero plus V zero T plus one-half a T squared or some variation on that there's equations like this V is equal to VZ plus 80 that's velocity is equal to the initial velocity plus the acceleration times the time we had on equations like this the final velocity or excuse me the the the average velocity is equal to the initial velocity plus the final velocity over two and there's another equation that said V squared is v-0 squared plus two a delta X and you might be tempted to take some of these equations and start trying to tackle this problem because these are the physics equations for accelerated motion the problem is these equations all deal with constant acceleration and that is not the case here look at this acceleration is a function of time so the acceleration is not constant these equations aren't going to help us but what will help us is understanding that velocity is the derivative of the position DX DT and acceleration is the derivative of the velocity DV DT those principles always apply so with that in mind let's solve the problem let's make a just a quick sketch the acceleration is going to look something like this acceleration as a function of time it's going to be some kind of parabola and we want to go from 0 to 7 and we're trying to find this area under the graph from 0 to 7 that will be the change in velocity the area under the acceleration graph is the change in velocity and if it starts at rest the change in velocity will be how fast it ends up moving so we solve this the change in velocity will be that area that's the integral from 0 to 7 of a DT so we just need to find the antiderivative of function a so what function has 1 point 2 T squared as its derivative well that's going to be 1.2 times T cubed over 3 and then we evaluate that from 0 to 7 and that means we put in the value 7 in for T and get an answer and value 0 in 40 and get an answer and subtract so it's going to be one point two times seven cubed over three minus and you don't even need to write this because this one is one point two times zero cubed over three you can leave that off that clearly is all zero so you just have one point two times seven cubed over three and you do the calculation and it comes out to one hundred and thirty seven point two and that's your answer that could be in meters per second or whatever unit we might be measuring in but one hundred and thirty seven point two is the exact area under this curve and the area under the acceleration graph is the change in velocity and let me just add that for a rocket this is reasonable a rocket would be expected to behave with a non constant acceleration because as a rocket is flying along through space remember how does a rocket propellant it blasts all this fuel out the back and so as it flies its mass decreases and as its mass decreases its acceleration increases so the acceleration isn't constant and there are plenty of other cases where acceleration isn't constant but we can still solve these problems although now we need calculus because we can't use the constant acceleration equations from physics calculus though comes to the rescue and allows us to find exact answers to problems like this

Channel: Derek Owens

Views: 431,970

Rating: 4.8408246 out of 5

Keywords: Calculus Ab, Derek Owens, Fundamental Theorem, Integral, Ap Calculus Ab, distance learning, homeschool math, Calc, Differentiation, Calculus, Integrating, Calc Ab, Integration, Math, homeschool, Integrate

Id: 9bJ2Z1jbIAE

Length: 10min 20sec (620 seconds)

Published: Mon Jan 12 2009

Reddit Comments

[u/Ob101010]

Just seems like were calculating an area and [waves magic wand] thats the answer. Id like to understand how calculating an area means solving this problem.

[u/nashef]

When you take the derivative of a function, the value of DX/day at each point is the slope of the line tangent to the function. When you integrate, the value of the integral is the area under the function. Those are the definitions.

It happens to turn out that they are inverses of each other, in a way.