- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Definite integrals properties review
Negative definite integrals
We learned that definite integrals give us the area under the curve and above the x-axis. But what if the curve itself is below the x-axis? In this case, the definite integral is still related to area, but it's negative. See how this works and get some intuition for why this is so.
Want to join the conversation?
- what happen when a is greater than b?(14 votes)
- The function will become negative because you are going in the reverse order.(8 votes)
- How the unit of the area is the only meter?
shouldn't it be Meter squared? at 3.40(1 vote)
- We're multiplying velocity, measured in meters per second (m/s), and time, measured in seconds (s), so the resulting unit will be
(m/s) ∙ s = m ∙ s/s = m ∙ 1 = m (meters).(7 votes)
- This question has always bugged me: which is greater: -8m/s or 4m/s? Or does it not make sense to ask this question at all?(1 vote)
- I think the difference between speed and velocity are key here.
Speed is pure numbers. There can be no negative speed. if a car is moving to the right at 4 m/s and another os moving left at 8 m/s, the left one may be moving in the negative direction, but speed doesn't account for that.
Velocity DOES take into account direction. That being said, it can have parts that are negative and parts that are positive. for instance if you had a space ship moving tot he right and down, right is the positive x directiona dn down is the negative y direction So it is very hard to compare things like that. It all depends on the question you are asking(4 votes)
- what if a=-6; b=-2; integral(f(x)=6), what would be the area?+24 or -24?(1 vote)
- If you're integrating from -6 to -2, you're taking the positive area because -6 is less than -2. f(x) = 6 is always above the x-axis, so this means that your area will be positive, as you're taking the integral in the normal direction of a function that has a positive area.(2 votes)
- at5:05, if a is greater than b, what's the difference and why?(1 vote)
- If a becomes greater than b, two possibilities come into play:
1. If the function is strictly below the x axis, the area will be negative. But, as your bounds are going from a higher number to lower number, on reversing them, a negative sign appears which negates the sign of the area, hence, giving a positive answer.
2. If the function is above the x axis, the area is positive. But, as we're going from a higher bound to a lower one, on reversing it, a negative sign comes and makes the area negative.(2 votes)
- [Instructor] We've already thought about what a definite integral means. If I'm taking the definite integral from a to b of f of x dx, I can just view that as the area below my function f. So if this is my y-axis, this is my x-axis, and y is equal to f of x, so something like that. Y is equal to f of x. And if this is a and if this is b, I could just view this expression as being equal to this area. But what if my function was not above the x-axis? What if it was below the x-axis? So these are going to be equivalent. Let's say, let me just draw that scenario. So let me draw a scenario where that's my x-axis, that is my y-axis. And let's say I have, let's say I have a function that looks like that. So that is y is equal to g of x. And let's say that this right over here is a, and this right over here is b. And let's say that this area right over here is equal to five. Well, if I were to ask you what is the definite integral from a to b of g of x dx, what do you think it is going to be? Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x-axis. You might be tempted to say, hey, this is just going to be equal to five. But you have to be very careful. Because if you're looking at the area above your curve and below your x-axis, versus below your curve and above the x-axis, this definite integral is actually going to be the negative of the area. Now, we'll see later on why this will work out nicely with a whole set of integration properties. But if you want to get some intuition for it, let's just think about velocity versus time graphs. So, if I, in my horizontal axis, that is time. My vertical axis, this is velocity. And velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds. And actually, I'm going to do two scenarios here. So let's say that I have a first velocity time graph. Let's just call it v one of t, which is equal to three. And it would be three meters per second, so one, two, three. So it would look like that. That is v one of t. And if I were to look at the definite integral going from time equals one to time equals five of v sub one of t dt, what would this be equal to? Well, here my function is above my t-axis. So I'll just go from one to five, which will be around there. And I could just think about the area here, and this area is pretty easy to calculate. It's going to be three meters per second times four seconds. That's my change in time. And so this is going to be 12 meters. And so this is going to be equal to 12. And one way to conceptualize this is this gives us our change in position. If my velocity is three meters per second, and since it's positive you can conceptualize that as it's going to the right at three meters per second. What is my change in position? Well, I would have gone 12 meters to the right. And you don't need calculus to figure that out. Three meters per second times four seconds would be 12 meters. But what if it were the other way around? What if I had another velocity function, ;et's call that v sub two of t, that is equal to negative two meters per second? And it's just a constant negative two meters per second. So this is v sub two of t right over here. What would or what should the definite integral from one to five of v sub two of t be, dt be equal to? Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left, as opposed to to the right. And so we can just look at this area right over here. When if you just look at it as the rectangle, it's going to be two times four, which is equal to eight. But you have to be very careful. Since it is below my horizontal axis and above my function, this is going to be negative. And this should make a lot of sense. If I'm going two meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds, then my change in position is going to be negative eight meters. I would have moved eight meters to the left, if we say the convention is negative means to the left. So the big takeaway is, if it's below your function and above the horizontal axis, the definite integral, and if your a is less than b, then your definite integral is going to be positive. If your a is less than b, but your function over that interval is below the horizontal axis, then your definite integral is going to be negative. And in the future, we'll also look at definite integrals that are a mix of both, but that's a little bit more complicated.