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Current time:0:00Total duration:9:45

Functions defined by integrals: challenge problem

Video transcript

so right over here we have the graph of the function f and we're assuming that f is a function of T our horizontal axis here is the T axis so that is f of T lowercase F of T and now let's just let's define another function let's call it capital F of and it's not going to be a function of T it's going to be a function of X so capital f of X is equal to we're going to define it as the definite integral between T is equal to negative 5 and T is equal to X actually let me make those X's in the same color that so they really stand out so f of X is equal to the definite integral between T is equal to negative 5 and T is equal to X of F of T F of T DT and if I had my druthers I probably wouldn't use capital F and lowercase F I would use like a G or capital G just so that when I say F you don't get confused but I'm going to try my best to say lowercase F of T capital f of X so this is how we're going to define the function capital f of X definite integral between T is equal to negative 5 and X of F of T DT now given this definition what I want to think about is that what X values does capital f of X does capital f of X equal 0 so let me write that down at what X values does this is this I guess you could say this equation at what X values is that is this equation true and I encourage you to pause this video right now and try to think about it on your own and then we can work through it together so I'm assuming you've had a go at it so let's just think about what this what this function capital f of X is really talking about well it's the one way to think about it is it's the area below between negative 5 between T equals negative 5 and T equals x that is below the function f of T and above the T axis and if the area is the other way around if it's below the T axis and above the function it's going to be negative area so we're looking at T equals negative five which is this you could say this boundary right over here that is T is equal to negative five and if you put pick an x-value let's just say x were negative two so if that is your x value right over here capital f would describe would describe this area in this area it would be a negative area because here the function is below the T axis so for example capital f of negative 2 would be negative now what x-values makes this zero so one might jump out at you well if we just if we made if we put X if we put X right at negative five right at negative five then there's no width width here there's not going to be any area so f of F of negative five F of negative capital F of negative five that should be clear here capital F of negative five which is equal to the definite integral between t is equal to negative five and T is equal to negative five of F of T DT F of D DT well you have the you have the you have the same boundaries here so this is going to be zero once again this area you have no width to this area so this is going to be equal to 0 so we can list x equals negative five as being one of the points one of the x-values that makes capital f of x equals zero but let's see if we can find more of them so let's see if we were to let's see if we were to go let me erase this right over here okay so this was we're starting at t equals negative five now as X gets larger and larger and larger when X is equal to negative three our area so let's see this area right over here this area right over here is is going to be so that is let's see this we have it's to going from negative 5 to negative 3 is 2 so this distance right over here is 2 this height right over here is 4 so this area right over here is 2 times 4 times 1/2 which is going to be 4 and since it's above the function and below the T axis we'll write this as a negative we'll write that as a negative 4 and now so let's see so we're starting when X is equal to 5 uf capital f of X is equal to 0 then as you get further and further out you're getting more and more negative values and it's more and more negative values but I just picked out this point right over here because this seems like a point of transition of the function and then we have this region which is just going to add more negative area to capital f of X as X gets larger and larger and larger and this looks like a really a quarter circle this is it has a 1/4 circle of radius 4 and now we have another quarter circle of radius 4 until we get to x equals 5 and this is actually going to be positive area because here our function is above the T axis so when we've gotten this far so all the way I'm thinking about it my X is my a Kuenn capital f of X equal negative 5 is zero and now our area as X becomes larger and larger larger our area becomes negative negative negative negative now it becomes less negative because we're starting to add we're starting to add so for example if X is equal to 2 if X is equal to 2 we would be looking at this positive but we still have all this huge negative area to overcome so we're still a negative territory but the more positive we add we're going to become less negative and you go all the way to x equals positive 5 and this positive area this quarter circle right over here positive area is going to exactly offset this quarter circle of negative area you don't have to think about what that area is although you can obviously figure it out with PI R squared and whatever else and so now we just have to keep adding more positive area to offset this negative for so how do we do that well the height here the height right over here is for the height right over here is four so if we if we can add our rectangle that is one wide and four high that's positive area of four which this has right over here so this is plus four it's going to offset this negative four and so we go all the way to X is equal to six when X is equal to six capital f of X is going to be equal to zero let's write that down so capital F capital f of capital f of positive six of positive six which is going to be equal to the definite integral between negative 5 and positive 6 positive 6 of F of T F of T DT well we can break this up and we already went through it I'm just going to going to make sure that we really understood what was going on this is equal to and I'll just do it all in one color now this is equal to actual I'll do it in these colors I did here this is equal to the integral the integral between negative 5 and negative 3 of F of T DT plus the integral plus the integral between negative 3 and 1 of F of T DT plus the integral between 1 and 5 of F of T DT that describes this right over here and then finally plus plus the definite integral between 5 & 6 of F of T DT F of T DT and this describes this negative area this describes this positive area these two net out to be 0 and then this area we already figured out is or this definite integral I should say is negative for this is negative for and this one right over here is positive for and so they net out and this of course is equal to zero which we wanted to figure out once again how did I do that well I said okay clearly when X is equal to negative five you have no area and then I just kept increasing kept increasing X's to larger and larger values I could have gone actually the other way and I would have had just more and more positive values and there but nothing to offset it to get it back to zero but as we increased X above negative five the capital f of X the area is more and more and more and more and more negative but then it becomes then we start adding positive value to it to offset the negative and we fully offset the negative at X is equal to 6