Proving the Fundamental Theorem of Calculus Part 2

Key Takeaways

in the previous video we looked at the fundamental theorem of calculus part one and we proved it now we're going to prove the part two the part most people use more commonly the one that allows us to do something like uh if we have x squared dx we can do one third x cubed and it goes from zero to one we can go from zero to one and we get the area under the curve of a parabola from zero to one is one third it's what allows us to do that so let's look at how to prove it first let's down let's write on the formal statement so the formal statement says we have to put some conditions if f continuous on a b closed interval a b then integral a to b of f of x dx equals f of b minus f uh let's write it over here minus f of a um where big f is any antiderivative of small x and that means that how f of x and f of x are related is that f prime of big f prime is equal to little f so we know that let's try to prove it so the first thing we're going to look at during the proof is that we're going to define a function g of x which is going to be defined by integral from a to x of f of t dt like we did in the previous fundamental theorems of calculus so now we're going to say that any other antiderivative so we know that g we know that g prime of x equals f of x so we know that uh g is an antiderivative of f so what else do we know if f if big f of x is any other anti-derivative of uh small f that means that g and f have to differ by only a constant right because if we have graphs if we have this graph and the derivative is something then if we have this graph that has the same derivative because it's just moved up and down by a constant so because of that we see that f big f of x equals g of x plus some constant c so that's going to be an important note later on so now we're going to do is we're going to take g of a and we're going to put that so remember g of a is going to be from a to a of f of t dt and that's going to give us 0 because it's going from the same point to the same point so if we're going from the same point same point and g of a remember g of x was defined as uh or rather f of x was defined as g of x plus c so if we do f of b minus f of a we're going to get g of b plus c minus uh g of a plus c so let's write that up here so you can see so we're going to get g of b plus c minus g of a plus c and put the brackets in the appropriate places so we can see the c's are going to cancel out it's going to be g of b minus g of a but we know g of a is zero so this is just equal to g of b and that is just going to be equal to a to b of f of t dt okay so what have we learned from this result so the full result the full result says that big f of b minus big f of a and f big f was just any antiderivative remember that equals the integral from a to b of f of t dt and that is exactly what the part two of the fundamental theorem of calculus says