- 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
Sal uses a graph to explain why we can take a constant out of a definite integral.
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- I still don't intuitively understand the connection between definite integral and derivatives; area under the curve and rate of a function. I believe Sal has explained that somewhere, but is there one more specific video on that topic or should I just watch them all?(35 votes)
- I had a similar confusion.
Then I read this article. It cleared my mind, I hope it clear yours too.
- at1:52, Sal says "a positive c greater than zero". but shouldn't it be "greater than one" since this is a vertical stretch (not a compression, from the way it was drawn)? thanks!(7 votes)
- Sal said he chose to draw it with c = 3, just because he had to use some value. But what he showed will work for any value of c > 0. He says the line may scaled up or down depending on c.(6 votes)
- I understand the property, but when i think about geometrical scalar factors, in those problems, the factor would increase both dimensions.
Unless, when dealing those scalars, each dimension is only receiving half of the increase, as oppose to this where the one dimensions receives all. For example, if c were 2, a geometric shape would actually only increase in each dimension by 1.5.
Is this right?(4 votes)
- In geometry the scalar factor would apply to all the dimensions. But we are not dealing with an equation for a geometric figure here. We are applying the scalar to a function. Visually f(x) is just a wavy line. So there are not "both dimensions" to scale. The C is only affecting the f(x)/y-values by moving the line up or down. The integral is an operation telling you to find the area under that line between a and b. This is where you begin to see a shape because you draw in the boundaries (called limits of integration) to help you visualize the area underneath your line. But in actuality, the function you're working with is just a line and never connects to itself to encloses an area like a geometric shape would. If it did, it would crease to be a function (fails the vertical line test). Hope this helps.(6 votes)
- What if the function is dilated in the x direction?
e.g. how would we find the area from x=a to x=b of the function f(c*x) ?(6 votes)
- Good question!
The integral (or net area) from x=a to x=b of f(c*x) dx is the same as (1/c) times the integral of x=ca to x=cb of f(x) dx.
(This comes from using the u-substitution u=cx which gives du=c dx, expressing everything in terms of u and du, changing the limits of integration accordingly, and then simply replacing all the u’s by x’s in the end.)(2 votes)
- We have studied that the integral of x^n is x^n+1/n+1. Can you elaborate?(2 votes)
- Remember the power rule? The derivative of x^n is nx^(n-1). You are integrating x^n. Since you want the exponent to be n after you differentiate the integral, take a look at the power rule. The exponent will be one lower when you differentiate. So, you need to take it one up when you are integrating. That gives x^(n+1). However, that is not the answer. From the power rule, the exponent also "comes down" to be a coefficient. You want a coefficient of one after you differentiate the integral, so you need to put 1/(n+1) in front. Your final answer for the integral of x^n: (x^(n+1))/(n+1)+C.
Note: In your question, you forgot to put +C. That term is very important in indefinite integrals.(8 votes)
- I DO NOT understand how does doing integration help us to get the area under or below or whatever of a curve...........please someone help me understand the theory behind it....(4 votes)
- Have you studied the Riemann section? It should instill a good level of intuition for finding the area to a function that is non-linear:
- what if c is being added to the function (f(x)+c) but the limits stay the same?(3 votes)
- Then c(b-a) will be added to the area under f(x) to get the area of f(x)+c(4 votes)
- If integration gives the area under a curve, will it be always positive?(2 votes)
- Not always. If the area you're integrating through is below the x-axis, you'll get a negative answer. Simple example is integration y=-1. The integral is -x. Integrating this between x=0 and 2, you'll get an area of -2.(5 votes)
- Why does y=f(x) graph and y=c*f(x) look different in shape?
Is there a reason or is it suppose to look the same but Sal simply made a small drawing error?(1 vote)
- When you multiply a function by a constant, you're just ''stretching'' it ''tightening'' it (above or below the x axis, depending on the sign of
- Can anyone tell me short way of calculating definite integral if upper limit is a positive number and the lower limit is zero.(2 votes)
- [Voiceover] We've already seen and you're probably getting tired of me pointing it out repeatedly, that this yellow area right over here, this area under the curve y is equal to f of x and above the positive x-axis or I guess I can say just above the x-axis between x equals a and x equals b, that we can denote this area right over here as the definite integral of from a to b of f of x dx. Now what I want to explore in this video and it'll come up with kind of an answer that you probably could have guessed on your own, but at least get an intuition for it, is that I want to start thinking about the area under the curve that's a scaled version of f of x. Let's say it's y is equal to c times f of x. Y is equal to some number times f of x, so it's scaling f of x. And so I want this to be kind of some arbitrary number, but just to help me visualize, you know I have to draw something so I'm just gonna kind of in my head let's just pretend the c is a three for visualization purposes. So it's going to be three times, so instead of one, instead of this far right over here it's going to be about this far. For right over here, instead of this far right over here it's going to be that and another right over there. And then instead of it's going to be about there. And then instead of it being like that it's going to be one, two and then three, right around there. So I'm starting to get a sense of what this curve is going to look like, a scaled version of f of x. And at least what I'm drawing is pretty close to three times f of x, but just to give you an idea is going to look something like, and let's see over here if this distance, do a second one, a third one, is gonna be up here. It's gonna look something like this. It's gonna look something like that. So this is a scaled version and the scale I did right here I assumed a positive c greater than zero, but this is just for visualization purposes. Now what do we think the area under this curve is going to be between a and b? So what do we think this area right over here is going to be? Now we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well one way to think about it is we just scaled the vertical dimension up by c, so one way that you could reason it is if I'm finding the area of something, if I have the area of a rectangle and I have the vertical dimension is let's say I don't want to use those same letters over and over again. Well let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area is going to be alpha times beta. Now if I scale up the vertical dimension by c, so instead of alpha this is c times alpha and this is, the width is beta, if I scale up the vertical dimension by c so this is now c times alpha, what's the area going to be? Well it's going to be c alpha times beta, or another way to think of it, when I scale one of the dimensions by c I take my old area and I scale up my old area up by c. And that's what we're doing, we're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now obviously that changes as our x changes, but when you think back to the Reimann sums the f of x was what gave us the height of our rectangles. We're now scaling up the height or scaling I should say because we might be scaling down depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c you're gonna scale the area by c. So this right over here, the integral, let me just rewrite it. The integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in the same color. We're going to take the area under the curve f of x from a to b f of x dx and we're just going to scale it up by this c. So you might say, "Okay maybe I could have felt "that was, you know, if I have a c inside the integral "now I can take the c out of the integral", and once again this is not a rigorous proof based on the definition of the definite integral, but it hopefully gives you a little bit of intuition why you can do this. If you scale up the function, you're essentially scaling up the vertical dimension, so the area under this is going to just be a scaled up version of the area under the original function f of x. And once again really, really, really useful property of definite integrals that's going to help us solve a bunch of definite integrals. And kind of clarify what we're even doing with them.