 Welcome back everyone. I want to do some examples now using the fundamental theorem of calculus part one to calculate the derivatives of integral functions We saw in the last video the statement of the fundamental theorem calculus part one and also its proof If you missed that, please click the link To take a look at it But just as a quick summary if we have a function g of x which is defined as an area function It's an integral function that is g is defined to be the integral of F of some function f as we range from x to a right here This would capture the area under f of t from a to x so g is this area function We saw last time that if you take the derivative of g that this is equal to f And some alternative notation right here if we take the derivative with respect to x of the function area function The integral from a to x of f of t dt This is none other than just f of x right there So taking the derivative of the integral gives you back the original function It's like taking f of x plus seven minus seven these This tells us that derivatives and integrals are inverse operations of each other and you can take this right here as a rule of differentiation just like the power rule the chain rule the product rule if you take the derivative of e to the x You get e to the x if you take the derivative of x cubed you get 3x squared We can accept this as a rule of differentiation that we take the derivative of an area function or aka an integral function You get back the integrand the original function So let's see an example of this you're gonna like these exercises This is pretty nice If we want to find the derivative of this area function g of x which is defined as the integral from 0 to x of The square root of 1 plus t squared dt. Well the derivative of G so g prime we are going to take the derivative with respect to x of this integral 0 to x square root of 1 plus t squared dt By the fundamental theorem of calculus part 1 or what we might write down as f tc 1 By the fundamental theorem calculus part 1 when you take the derivative of an integral function This will just be The integrand the square root of 1 plus x squared do make sure to switch the variable there If you typed in a t until like the computer it would recognize that isn't correct because the variable of g here is x so as x is changed is what we're trying to measure And so that's it by the fundamental theorem of calculus to calculate this derivative of the integral We just keep back the original function. There's really much much more to it except for Erasing the integral part and I also want to make mention that when you use the fundamental theorem of calculus part 1 Two things you should notice one there should be a There should be a constant as the lower limit of the integral What is that constant? It really doesn't matter 0 negative 17 negative pi 300,000 5,000 280 whatever that constant is it doesn't really matter, but you do need a constant at the bottom of your integral Otherwise f tc 1 doesn't apply The other thing you need is you need a variable at the very top in this case. You see just an x It needs to be a variable at the top and a constant at the bottom When you take the if you take the derivative Then you'll just get back the function that we see right here If it doesn't match that format you can't use the fundamental theorem calculus part 1 to take the integrative Immediately it turns out that you can use properties of integration to compensate for those and I'll show you some examples of those In the next video take a look for those Bye