in this video we're going to talk about the first part of the fundamental theorem of calculus so if g of x is equal to the definite integral of f of t from a to x then g prime of x is going to equal f of x so if g is the antiderivative of f then the derivative of g will equal f the derivative of the antiderivative will give you the original function another way in which you can express the second part of the i mean the first part of the fundamental theorem of calculus is you can express it this way the derivative of the integral from a to x of f of t dt is going to equal f of x and for the most part you're replacing t with x but that's not always the case as some examples will illustrate but that's the basic idea behind the first part of the fundamental theorem of calculus some textbooks may call it the second part of the fundamental theorem of calculus but regardless of what they call it the principle still remains the same now let's work on an example problem so let's find the derivative with respect to x of the integral from zero to x of the function the square root of t squared plus 4 dt now it's going to be difficult to integrate this function and then differentiate it so there has to be an easier way perhaps you realize that the answer is going to be f of x which is just the square root of x squared plus 4. but there's a process that you can employ to get the answer and we're going to go over that right now so f of t is going to be the function square root t squared plus 4. so what we now have is just the integration of f of t dt from 0 to x now capital f is the antiderivative of lowercase f so let's find the antiderivative it's going to be capital f of t evaluated from 0 to x and that's equal to f of x minus f of zero now the derivative of f of x that is capital f of x is going to give us lower case f of x and f of 0 is a constant the derivative of a constant is 0. so this is equal to f of x now if f of t is equal to the square root of t squared plus 4 then f of x is going to be the square root of x squared plus 4. and so this is the final answer to the entire problem now let's try another problem so let's say it's going to be the antiderivative or the integral of let's say the square root of t to the third plus five from x to four go ahead and try that feel free to pause the video so let's define f of t as being the square root of t to the third plus five and so we have this expression we can replace this with f of t the antiderivative of lowercase f we know it's capital f evaluated from x to four so then this is going to be f of 4 minus f of x now f of 4 is a constant and so its derivative will be 0. now the derivative of capital f is going to be lower case f so notice that the answer is not just f of x anymore it's negative f of x if x is in the bottom if x was on top then it would become f of x so the final answer is going to be negative square root x cubed plus five and so the answer can change so it's not always just f of x but if you go through this process you can always get the right answer here's another example so instead of x let's say it's x squared we're going to have the cube root actually let's use the square root of t to the third minus 4 dt so go ahead and try it what do you think the answer is going to be if you had to guess it well here's what you need to do if you want to do it the fast way the first thing you need to do is replace t with what you see here in this case x squared so it's going to be x squared raised to the third power minus four and then you got to multiply by the derivative of x squared based on the chain rule if it was just x the derivative of x is one so it wasn't important then but now it is so the final answer is going to be 2x times the square root of x to the 6 minus 4. now let's confirm it so i'm just going to write this answer at the bottom so let's go through that process that we've been going through so first let's define f of t so f of t is going to be the square root of t to the third minus four and so we have this expression f of t dt and the anti-derivative of lowercase f as always that's not going to change it's going to be capital f evaluated from 5 to x squared and so this is going to be f of x squared that's a terrible looking 2 minus f of 5. so what is the derivative of f of x squared well we need to use the chain rule so we need to differentiate the outside part of the function the derivative of capital f is lowercase f then we need to keep the stuff on the inside the same and then we need to find the derivative of the inside function which is 2x now this is a constant and the derivative of any constant will be zero so now f of t is equal to this so what is f of x squared we need to replace t with x squared so it's going to be x squared raised to the third power minus four so then that becomes the square root of x to the sixth minus four and then times two x so as you can see these two are the same let's try one more example and i recommend pausing the video and working on it if you can get this one right then you're okay with this topic so we're going to integrate the function from x squared to x cubed and let's say it's the square root of t to the fourth power minus 2. so go ahead and try so let's do it the easy way first we're going to plug in the top part of the integral the upper limit so it's going to be the square root of replace t with x cubed so x cubed raised to the fourth power minus two and then multiply by the derivative of x cubed so that's going to be 3x squared and then for the bottom don't forget to subtract and let's replace t with x squared it's going to be x squared raised to the 4th power minus 2 and the derivative of x squared is 2x so the final answer which i'm going to write here is going to be 3x squared square root x to the twelfth minus two and then minus two x square root x to the eighth minus two and so that's the fast way of getting the answer but let's show our work so once again we're going to define f of t as being the square root of t to the fourth minus two so now we have f of t dt and the antiderivative of lowercase f is capital f evaluated from x squared to x cubed and then this is going to be f of x cubed minus f of x squared and the derivative of the outside function capital f is lower case f keep the inside the same and then differentiate the inside function which is 3x squared and then it's going to be minus f of x squared times the derivative of x squared which is 2x so now using this function let's plug in x cubed so it's going to be the square root of x to the third raised to the fourth minus two and we can move this to the front and then it's gonna be minus two x square root x squared raised to the fourth minus two so you can see we're going to get the same final answer and so that's how you can apply the fundamental theorem of calculus part one if your textbook defines it as part one you