Express the following limit as a definite integral on the interval [ 0 , 10 ] [0,10] . lim ⁡ n → ∞ ∑ k = 1 n ( x k ∗ − 3 ) 2 Δ x \lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta x

∫ 0 10 ⁣ ( x − 3 ) 2 d x \int_0^{10}\!\left(x-3\right)^2\,dx

∫ 10 0 ⁣ ( x − 3 ) 2 d x \int_{10}^0\!\left(x-3\right)^2\,dx

∫ 0 ∞ ⁣ ( x − 3 ) 2 d x \int_0^{\infty}\!\left(x-3\right)^2\,dx

∫ ∞ 0 ⁣ ( x − 3 ) 2 d x \int_{\infty}^0\!\left(x-3\right)^2\,dx

Given the following definite integral of the function f ( x ) = 3 x 2 − 2 x f\left(x\right)=3x^2-2x , write the simplified integral: − ∫ 4 0 ⁣ f ( x ) d x -\int_4^0\!f\left(x\right)\,dx − ∫ 4 0 f ( x ) d x

3 ∫ 0 4 ⁣ x 2 d x − 2 ∫ 0 4 ⁣ x d x 3\int_0^4\!x^2\,dx-2\int_0^4\!x\,dx

2 ∫ 4 0 ⁣ x d x − 3 ∫ 4 0 ⁣ x 2 d x 2\int_4^0\!x\,dx-3\int_4^0\!x^2\,dx

3 ∫ 4 0 ⁣ x 2 d x − 2 ∫ 4 0 ⁣ x d x 3\int_4^0\!x^2\,dx-2\int_4^0\!x\,dx

2 ∫ 0 4 ⁣ x d x − 3 ∫ 0 4 ⁣ x 2 d x 2\int_0^4\!x\,dx-3\int_0^4\!x^2\,dx

Write the two definite integrals subtracted below as a single integral. ∫ 1 6 ⁣ x 2 − 5 x d x − ∫ 10 6 ⁣ x 2 − 5 x d x \int_1^6\!\sqrt{x^2-5x}\,dx-\int_{10}^6\!\sqrt{x^2-5x}\,dx

∫ 1 10 ⁣ x 2 − 5 x d x \int_1^{10}\!\sqrt{x^2-5x}\,dx

∫ 10 1 ⁣ x 2 − 5 x d x \int_{10}^1\!\sqrt{x^2-5x}\,dx

∫ 6 9 ⁣ x 2 − 5 x d x \int_6^9\!\sqrt{x^2-5x}\,dx

∫ 9 6 ⁣ x 2 − 5 x d x \int_9^6\!\sqrt{x^2-5x}\,dx

8. Definite Integrals

Introduction to Definite Integrals

8. Definite Integrals

Introduction to Definite Integrals: Study with Video Lessons, Practice Problems & Examples

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To find the area under a curve, we use the concept of the definite integral, which is the limit of Riemann sums as the number of rectangles approaches infinity. The definite integral is expressed as $\int f_{a} b_{x}$ . Key rules include the sum and difference rule, constant multiple rule, order of integration rule, zero width interval rule, and additivity rule, which help simplify calculations and understand the properties of integrals.

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Definition of the Definite Integral

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One of our big focuses in recent videos has been estimating the area under the curve of a function. And recall that our main strategy for doing this is filling this function with rectangles and calculating the area of each rectangle and adding it up. But again, this only gives you an approximation or an estimation for how much area there is. But what if I told you there was a way to find a truly accurate answer for how much area we had under the curve of a function? Well, it turns out there is a way of doing this, and right at first, it might seem a bit complicated.

But pay close attention because I'm going to walk you through how to set this up, and hopefully, this is going to be very clear. So let's just go ahead and get right into things. Now I want you to recall that whenever we're estimating this area, the more subintervals or the more rectangles that we use is going to give us a more accurate answer to this Riemann sum, which tells us the approximate area under the curve. Ӂhe question becomes, if more rectangles means more accurate, then how exactly could we get a completely accurate answer for the area under a function? Well, the way that we could do this is by recognizing I need to take this function and fill it up with infinite amounts of rectangles.

So if the number of subintervals or rectangles I have approaches infinity, this will give the true area under the curve of the function. Now when we have the true area underneath our curve, we call this the definite integral. So we would say that the true area under here is what is the definite integral for this function. Now the definite integral is going to look something like this. So first off, we have this setup for our definite integral.

Now I want you to notice something about this left side of the equation here. Notice this looks very similar to what we've already seen when it comes to dealing with Riemann sums. The main difference though, is that we have this limit out in front. This limit is going to approach infinity, because we need an infinite number of rectangles to fill underneath the curve of this function. So since we have this infinite number of rectangles, that means that we're dealing with a definite integral.

Now you may recall that we talked about indefinite integrals, and indefinite integrals had this little kind of squiggly line, and then we were integrating some function with respect to whatever variable we had inside that function. Well, the idea with definite integrals is very similar, except notice with a definite integral, we're now looking at two places. We're looking at these bounds on the x axis, which is why we need to use a squiggly line with these two values right there. So what the squiggly line is going to do is it's going to go from a to b, where a is the lower bound and b is the upper bound on the x axis. And this tells us where we need to find our area for our function.

So this is what we do when setting up a definite integral. But to make sure that we understand how this works, well, let's take a look at an example down here. So notice in this example, we're asked to express the following limit as a definite integral on the interval from zero to four. Now, to start off this problem, what I'm going to do is take a look at the function I'm dealing with, what this curve looks like, and think about how I can set this up as an integral. Now first off, notice we are given this summation right here.

This summation follows a similar pattern as a Riemann sum does, except notice we now have n approaching infinity. Since the number of rectangles or subintervals approaches infinity, we know that this is going to be a definite integral setup that we can do. So to set up this definite integral, I know that I'm going to have this integral sign that goes from the lower bound to the upper bound. I can see that the lower bound is going to start at zero, and the upper bound finishes at four. So we go from zero to four on our x axis.

So what that means is that we're looking at this point of zero, we're looking at this place of four on the x axis, and we're trying to estimate what this area would be underneath the curve of our function. So this is what we're trying to calculate, and this will be how you could set up the bounds on the definite integral right here. Now what I'm doing is I'm going to be integrating this function x plus one with respect to our variable x. So this would be the setup for our definite integral, and that is how you can set up problems like this where you're trying to find the area under the curve. Now this is not the final answer to the problem because notice we're also asked to find the exact area that it represents.

And to find the exact area, well, what I need to do is look over at my function here on the graph and see if there are any familiar shapes that form. Well, what I can do is take this and split it like so, like a graph here, and notice this gives me two familiar shapes, a triangle and a rectangle. So I need to find the area of this triangle, the area of this rectangle, and if I add them all up, it will give me the true area under the curve of the function. So let's do that. So the area one that we're dealing with is going to be this triangle.

One half base times height is the equation for the area. Now I can see that the base is going to be from zero to four on our x axis, so that's going to be four, the base, and the height goes from one all the way up to five on the y axis, and the distance from one to five would be four. So we have one half four times four, which gives us an area of eight. Now next, I need to find the area of this rectangle. The area for any rectangle is base times height.

I can see that the base goes from zero to four, and then I can see the height goes from zero to one on the y axis. So we have four times one, which is four. So our total area is going to be eight plus four, which is 12. So 12 would be the area under the curve of this function, and this is how we can set up the integral and find the true area. So this is how you can solve these types of problems where you need to set up definite integrals and you need to figure out what they represent.

And hopefully, by going through this example, it makes sense that the definite integral will give us the true area under the curve of our function. So this is the process for doing this, and you're going to see a lot of examples and a lot of applications for this definite integral. So I highly recommend you check out the practice and examples on this video, and also check out the future videos where we're going to dive more into how we can solve definite integrals and the various rules and problems that we'll see. See you in the next one.

Problem

Find $integral from 0 to 6 comma f of x d x$ given the graph of $y = f (x)$ .

Problem

Express the following limit as a definite integral on the interval $open bracket 0 comma 10 close bracket$ .
$\lim_{n \to \infty} \sum_{k = 1}^{n} {(x_{k}^{*} - 3)}^{2} Δ x$

$integral from 10 to 0 comma open paren x minus 3 close paren squared d x$

$integral from 0 to 10 comma open paren x minus 3 close paren squared d x$

$integral from 0 to infinity comma open paren x minus 3 close paren squared d x$

$integral from infinity to 0 comma open paren x minus 3 close paren squared d x$

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Basic Rules for Definite Integrals

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In the last video, you may recall that we learned about definite integrals. We talked about what a definite integral looks like and what it represents. Now in this video, we're going to see that there are some basic rules that apply to definite integrals. You may also recall indefinite integrals, which had some basic rules that apply to them. And it turns out some of these same rules also apply to definite integrals.

If this isn't familiar to you, don't sweat it because I'm going to walk you through it today in this video, and we're also going to see some specific rules that only apply to definite integrals. So without further ado, let's get into things. One of the first rules that we learned about for indefinite integrals is the sum and difference rule. And all the sum and difference rule says is that if you have the integral of two functions being added or subtracted, you can split that into two separate functions being integrated. So notice we have the integral of the first function, and then we have the integral of the second function.

When this happens, the bounds, a and b, will stay the same. So, a and b are going to be the same for these integrals. We'll go from a to b, our low bound to high bound. And then we'll also go from a to b on the second integral for the second function g(x). An example of this would be if we were integrating x plus six from negative two to five.

What I could do is split this to the integral of x and the integral of six separately, and the bounds would still go from negative two to five on both integrals. So this is the same rule that happens for indefinite integrals, and it also happens for definite integrals where the bounds do not change. So this would be the solution for our integral right here. Next, we need to look at the constant multiple rule. We also learned this for indefinite integrals, and it turns out the rule is the same for definite integrals.

This just says if you have a constant in front of your function, you can take that constant and pull it to the outside of the integral. And what we can do is keep the bounds the same. So this will still go from a to b when we do this rule. Notice that we have the integral from two to six of the function five x cubed.

Notice we have a five outside the function, which we can actually take that and pull it outside of the entire integral right here. Now what happens is we'll keep the same bounds, so we'll still go from two to six, and we're just going to have the integral of x cubed, which is going to be more simple to solve than when we had that five in there. So we can just take the integral of x cubed, then multiply it by the five, and that would give us the solution to this integral. As you can see, those rules are pretty straightforward since they're the same for indefinite and definite integrals. But things will get a little bit more complicated as we move on to these next rules.

The reason why these are a bit more complicated is that now we're focusing on the definite integrals like we were up here, except we need to focus on their bounds because the bounds of the definite integral are going to create some specific rules that we haven't seen before. The first example we're going to take a look at is the order of integration rule. The order of integration says that you can take the bounds b to a, and you can switch them and make them a to b. Notice that we have b to a here. We switch the bounds.

What happens is whenever you do this, you have to make your entire integral negative. An example of this would be like what we have here. We have the integral from four to one of x cubed. Now I don't like that I have this four in the bottom here, my bigger number. So what I'm gonna do is take that four and bring it to the top, then take this one and bring it to the bottom.

I'm just switching the order of the integral. Now we're going to be integrating the same function x cubed, so this doesn't change. But the only difference here is we have to make the entire result negative since we switched the bounds. So that would be the solution if we change the order of the bounds. Next, we'll take a look at the zero-width interval.

The zero-width interval rule states that if we have the same bounds, like a to a, what that's going to do is take whatever our function is, and it's going to make it zero when we integrate it. I'm assuming that f(a) exists. Now why is this the case? Why do we get zero when we integrate from a to a? Recall something about definite integrals.

Recall that these tell us the area underneath the curve of some function. And if I'm looking at a specific place, let's just say, a, and I'm integrating from a to a, notice we just get a straight line right there. We're not going from a to some other value b like we've seen before where we actually do have some area. We're just going from a to a, and there's going to be no area that we have in there, so that's why we get an area of zero. So that's the reason why when we integrate from one bound to the same bound, we just get a result of zero every time.

For example, let's say we had this polynomial right here that goes from three to three. Well, because it goes from one bound to the same bound, I can see that this whole result just turns out to be zero, and that would be the solution for this integral. The last rule we're going to take a look at is the additivity rule. This one you want to pay attention to because there are some things that need to apply for this rule to work. Notice in this case, we go from a to c of our first function, and then we go from c to b for the same function.

And when this happens, what we can do is we can actually write this as one integral. We can combine it. So if we're going from a to c, and then we're going from c to b, assuming that this c value is in between a and b, then what we can do is see that this whole_integral is just going to go from a to b of the same function f(x). So we're just going to have f(x) right here. And notice how we were able to take these two integrals split, and we were able to create it and combine it into one integral.

An example would be like what we have on the right side here. Notice in this example, we are given the integral from one to three, and we are also given the integral from three to seven. But notice it's the same function that we have here, five x to the fourth power. How exactly would we deal with this integral? We'll notice that we have this three right here, and three is between one and seven.

So because of that, we can write this as one integral that simply goes from one to seven. Now what's inside this integral is going to be our function that we have right here, five x to the fourth power, and all of that is going to be with respect to x. This is how we can deal with the additivity rule when it comes to these integrals. Something that I'm gonna mention as we move forward and take a look at more examples and practice for these problems that we've seen, what we're going to see is that multiple rules will be used together to find integrals of more complicated functions. So I highly encourage you to check out the practice and examples after this video to make sure we know how to solve these types of 문제들.

These are the basic rules that you can apply whenever you're dealing with definite integrals. See you in the next video.

Problem

Given the following definite integral of the function $f (x) = 3 x^{2} - 2 x$ , write the simplified integral:
$-\int_4^0\!f\left(x\right)\,dx$

$2 integral from 4 to 0 comma x d x minus 3 integral from 4 to 0 comma x squared d x$

$3 integral from 4 to 0 comma x squared d x minus 2 integral from 4 to 0 comma x d x$

$2 integral from 0 to 4 comma x d x minus 3 integral from 0 to 4 comma x squared d x$

$3 integral from 0 to 4 comma x squared d x minus 2 integral from 0 to 4 comma x d x$

Problem

Write the two definite integrals subtracted below as a single integral.
$\int_{1}^{6} ⁣ \sqrt{x^{2} - 5 x} d x - \int_{10}^{6} ⁣ \sqrt{x^{2} - 5 x} d x$

$\int_{10}^{1} ⁣ \sqrt{x^{2} - 5 x} d x$

$\int_{1}^{10} ⁣ \sqrt{x^{2} - 5 x} d x$

$\int_{6}^{9} ⁣ \sqrt{x^{2} - 5 x} d x$

$\int_{9}^{6} ⁣ \sqrt{x^{2} - 5 x} d x$

Problem

Evaluate the following definite integral.
$\int_{100}^{100} ⁣ \frac{\sin (2 x) - 100 x^{2}}{\sqrt{x^{5} + \tan (3 x - 6)}} d x$

Indeterminate

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Introduction to Definite Integrals: Study with Video Lessons, Practice Problems & Examples

Definition of the Definite Integral

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Find ∫06 ⁣f(x) dx\int_0^6\!f\left(x\right)\,dx given the graph of y=f(x)y=f\left(x\right).

Express the following limit as a definite integral on the interval [0,10][0,10].lim⁡n→∞∑k=1n(xk∗−3)2Δx\lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta x

Basic Rules for Definite Integrals

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Given the following definite integral of the function f(x)=3x2−2xf\left(x\right)=3x^2-2x, write the simplified integral:﻿−∫40 ⁣f(x) dx-\int_4^0\!f\left(x\right)\,dx−∫40​f(x)dx﻿

Write the two definite integrals subtracted below as a single integral.∫16 ⁣x2−5x dx−∫106 ⁣x2−5x dx\int_1^6\!\sqrt{x^2-5x}\,dx-\int_{10}^6\!\sqrt{x^2-5x}\,dx

Evaluate the following definite integral.∫100100 ⁣sin⁡(2x)−100x2x5+tan⁡(3x−6) dx\int_{100}^{100}\!\frac{\sin\left(2x\right)-100x^2}{\sqrt{x^5+\tan\left(3x-6\right)}}\,dx

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Find $integral from 0 to 6 comma f of x d x$ given the graph of $y = f (x)$ .

Express the following limit as a definite integral on the interval $open bracket 0 comma 10 close bracket$ .
$\lim_{n \to \infty} \sum_{k = 1}^{n} {(x_{k}^{*} - 3)}^{2} Δ x$

Given the following definite integral of the function $f (x) = 3 x^{2} - 2 x$ , write the simplified integral:
$-\int_4^0\!f\left(x\right)\,dx$

Write the two definite integrals subtracted below as a single integral.
$\int_{1}^{6} ⁣ \sqrt{x^{2} - 5 x} d x - \int_{10}^{6} ⁣ \sqrt{x^{2} - 5 x} d x$

Evaluate the following definite integral.
$\int_{100}^{100} ⁣ \frac{\sin (2 x) - 100 x^{2}}{\sqrt{x^{5} + \tan (3 x - 6)}} d x$