Class 12 math (India)
- Worked example: Breaking up the integral's interval
- Functions defined by integrals: switched interval
- Functions defined by definite integrals (accumulation functions)
- Functions defined by integrals: challenge problem
- Functions defined by integrals challenge
Functions defined by integrals: challenge problem
Solving for where a function defined by an integral equals 0. Created by Sal Khan.
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- Couldn't you also go from -7 to positive 5?(6 votes)
- Correct me if i'm wrong anyone but I believe we had to start from -5 as one of the conditions of the question. (Seen in top integral finding area from -5 to a value of x). Sorry if this wasn't explained very well(46 votes)
How does Sal know it's a quarter circle? I mean, if it were a written test I'd probably have to prove both purple areas are equal, and that involves proving they have the same "format". Is there a way to spot a part of a circle just by seeing it?
(I'm considering the existence of infinite ways for going from (1;1) to (5;4), ranging from a straight line to a 90 degrees part of a circle).(16 votes)
- The thing I noticed was that the radius at t = -3 was 4 and the radius from (-3, 0) to (1, 0) was also 4, then the radius from (1, 0) to (5, 0) is ALSO 4 and the radius at x = 5 is 4. A curve that has a
constant radiusis following a segment of a circle. That was enough justification in my opinion for assuming a quarter circle for the purposes of solving these exercises--this curve was looking like a piece of a circle, and had a constant radius at the points I checked. If you really wanted to check the curve at another point, the circle passes through or very close to (-1, -3.5). Using the distance formula, this gives a radius ≈ 4.03, which is very close to 4, and if you squint at the curve, the 3.5 is actually an over-estimate, but at this scale, it is difficult to give more decimal places. (In order to be a radius of 4, the coordinate would be ≈ 3.464 --an approximation of √12 , and it isn't possible to derive that much from the drawing.) The other quarter circle passes through or very close to (3, 3.5). This point also gives a radius of ≈ 4.03. A third checkpoint is (1.5, 2), which is also ≈ 4.03. For the purposes of evaluating these integrals, that is close enough.(12 votes)
- Could I integrate from -5 to -7 i.e. going left on the x-axis?(3 votes)
- Yes, that would work. However, some instructors don't like that and would mark off for it. If you reverse the position of the bounds, the answer will be the negative of the original form. Thus,
∫ f(x) from a to b = − ∫ f(x) from b to a(6 votes)
- Negative area is related to whether or not the function is below the x-axis, right? So even if I have negative xs-values on my function, the area between the function and the x-axis would be positive if the function was positive on that "negative x interval", right?(1 vote)
- The interpretation of an area below the x-axis depends on what is being represented by the function. For example, if the area represents something like profits and a negative area represents losses, then that would be subtracted. But, if the area represented, say, the area of a shape to cut a fabric, then the area below the x-axis would still be a physical area and be treated as a positive area. So, whether the area below the x-axis is treated as a positive or negative depends on the context and what the function/integral is representing.(7 votes)
- I am having a hard time understanding how "x" is not called "t" since he is explaining x values on the t axis. Couldnt the integral just be called from -5 to "t" of f(t)dt? Why call it x?(2 votes)
- Indeed he could have again used the variable
t, he decided to change the name of the variable to avoid confusion. while inside the integral
tis the integration variable, and since this is a definite integral, the integration variable will disappear once the integral is evaluated. To prevent confusion of a variable
tstill existing after the integral was evaluated, Sal decided to change the name of the variable to
x, to represent the variable once the integral was evaluated.(1 vote)
- How would you go about doing a problem like this algebraically without a graph?(2 votes)
- Why couldn't I have done the Integral between -5 to 1 + Integral between 1 to 6?(1 vote)
- Because you're not finding f(t)'s zeroes or anything like that: you're simply finding the zeroes of F(x). In this case, you want the total area of f(t) between itself and the x-axis to be nullified. Those two areas would indeed zero each other, but that won't answer the question (especially when not given an equation for f(t)), if that makes any sense. You need to find the values for F(x)'s zeroes without an equation, as Sal does in the video.(2 votes)
- what is integral from sin(-pi) from sin(pi)?(1 vote)
- It would be zero also, via symmetry of the sin function. The negative area between -pi and 0 is equal to the positive area between 0 and pi.
See https://www.desmos.com/calculator/hu5yauacj3(2 votes)
- At2:14, how do you figure out deciding that the area would be below f(t) and above the t-axis? Why can't it be, say above f(t) and below the t-axis?(1 vote)
- What about F(0) = 0? or F(1) = 0?(1 vote)
- That wouldn't equal zero. Because F(X) was defined as the integral from -5 to x. So Setting x as 0 or 1, you could look at the graph and see that from -5 to either of those values, has a lot of negative area between the function and the x axis(1 vote)
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.