Functions defined by integrals: challenge problem

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's 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 gonna definite 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 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'd probably wouldn't use capital f and lower case f, I would use like a g or capital G just so that when I say F, you don't get confused. Well I'm gonna try my best to say lower-case F of t, capital F of x. So this is how we're gonna 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 wanna think about is at what x values does capital F of x, does capital F of x equal 0? So we'll write that down. At what x values does, this is, is this, I guess you could say, this equation, at what x values 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 areas is the other way around, if it's below the t axis and above the function, it's gonna be negative area. So we're looking at t equals negative 5, which is this, you could say this boundary right over here. That is t is equal to negative 5. And if you put, pick an x value, let's just say x were negative 2, so if that is your x value right over here, capital F would describe, would describe this area, 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 a 0? 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 5. Right at negative 5. Then there's no width, width here. There's not going to be any area. So f of, f of negative 5. F of negative 5, F of negative capital F of negative 5, I should be clear here. Capital F of negative 5, which is equal to the definite integral between t is equal to negative 5 and t is equal to negative 5, of f of t dt. F of t dt. Well you have the, you have the, you, you have the same boundaries here. So this is going to be 0, 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 5 as being one of the points. One of the x values that makes capital F of x equal 0. But let's see if we can find more of them. Let's see if we were to go, let me erase this right over here. So, this was. We're starting at t equals negative 5. Now as x gets larger and larger and larger, when x is equal to negative 3, our area. So, let's see this area right over here, this area right over here is going to be, so that is, I see that we have, it's 2, we're going from negative 5 to negative 3. So the distance right over here is two. 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, that is you get further and further out, you're getting more and more negative values, that'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 quarter 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 x, when capital F of x equal negative 5 is, is 0. And now our area, as x becomes larger and larger, and 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 in negative territory. But the more positive we add, we're gonna become less negative. And you go all the way to x equals positive 5 and this positive area, this quarter circle right over here of positive area, is going to exactly offset this quarter circle of negative area. You don't even 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 4. So how do we do that? Well the height here, the height right over here is 4, the height right over here is 4 so if we, if we can add our rectangle that is one wide and 4 high. That's positive area of 4, which this has right over here. So this is plus 4. Is going to offset this negative 4. So we go all the way to x is equal to 6. When x is equal to 6, capital F of x is going to be equal to 0. Let's write that down. So capital F. Capital f of, capital f of positive 6, of positive 6, 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. We already went through it, I'm just gonna, going to make sure that we really understood what was going on. This is equal to, and I'll just to it all in one color now. This is equal to, actually I'll do it in these colors that 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 and 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 4. This is negative 4 and this one right over here is positive 4 and so they net out and this of course is equal to 0 which we wanted to figure out. Once again how did I do that? Well I said okay clearly when x is equal to negative 5 you have no area and then i just kept increasing, kept increasing xs to larger and larger values, I could have gone actually the other way. And, I would have had just more and more positive values there would have been nothing to offset it to get us back to 0. But, as we increased x above negative 5, the capital Fof x; the area, is 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.