WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to do some example problems for the different integral that we introduced in the last lesson.
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Let us get started.
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Approximate the following definite integral using mid points and the given number of sub intervals.
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This is a lot like the problems that we had before, when we are dealing with area.
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We want to just go ahead and start with approximation using midpoints.
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I think I will work in blue here.
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We have the integral from 1 to 6.
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Δx is equal to, we want 5 sub intervals.
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We have 6 - 1/5 which is equal to 1.
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Our Δx is equal to 1.
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We are going from 1, 2, 3, 4, 5, 6.
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This is 1, this is 6, we are going to be looking at these endpoints.
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These are going to be our x sub i.
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This is going to be x₁, this is going to be x₂, x₃, x₄, and x₅.
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1.5, 2.5, 3.5, 4.5, and 5.5, because we are approximating with midpoints.
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Let us go ahead and set up a little table here.
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Let me do it over here, actually.
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I have got my x sub i and I have my f(x sub i).
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I have got 1.5, 2.5, 3.5, 4.5, and 5.5.
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When I put these numbers into this function to find out what the y value is, I get 0.753, 1.283, 1.295, 1.012, and 0.6799.
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Therefore, the integral from 1 to 6 of x³ e ⁻x dx is going to be approximately equal to the sum as i goes from 1 to 5.
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5 points of the f(x sub i) Δx.
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That is just equal to Δx which is 1 × the sum,
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I’m sorry, I will write all of this out.
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This thing is nothing more than Δx × f(x₁) f(x₂) + f(x₃) + f(x₄) + f(x₅).
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Let us go ahead and be as explicit as possible.
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That is going to equal 1 × 0.753 + 1.283 + 1.295 + 1.012 + 0.6799.
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When I do all of that, I end up with 5.0229, that is my approximate integral using midpoints.
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We could have used right endpoints, we could have used left endpoints, 1, 2, 3, 4, 5.
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You are going to get numbers that are sort of similar.
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We decided to use midpoints for this particular problem.
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Just to show you what it actually looks like.
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Here is my list of x values, 1.5, 2.5, 3.5, 4.5.
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These are the actual values of f(x) carried out to a greater degree of precision.
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Our function happens to look like this.
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We were going from 1 to 6 midpoints, numbers.
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Midpoints give me all the numbers.
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What we found was the definite integral.
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In this particular case, between 1 and 6, the function happens to be positive, it is above the x axis.
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Therefore, this number that we got, the 5.02 something, it happens to be the area under the curve.
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It happens to be, it is the integral from 1 to 6 of this function using midpoints.
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An approximation of the integral, it just happens to correspond with the area under the curve.
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Express the following limit as a definite integral.
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Little practice using the actual symbol, very simple.
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The limit is an goes to infinity 1 to n, here is our function.
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This is going to be our integrand, this we are going to turn into dx instead of Δx.
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This is going to be our lower limit integration, this is going to be our upper limit of integration.
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We write this as the integral from π/3 to π cos² x/ x² dx.
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We are done, this is the definition, the limit of the sum.
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The summation symbol becomes the integral symbol.
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That is what is happening here.
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For the real problem, evaluate the following definite integral using the definition.
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This is going to be a bit of a process here.
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One of the reasons why we want to come up with quick ways of doing the integral was because
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we do not want to take this process that we are about to go through, for this problem in the next, every time we take an integral.
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This is going to equal the limit as n goes to infinity of the sum from 1 to n of the x sub i Δx.
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First, we find this, in other words we find an expression for that given the fact we have f(x) and we have -2 and 4.
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First, we find that.
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Second, once we find that, we form the sum.
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Second, we are going to evaluate the sum.
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When we evaluate the sum, you will end up with the function of n.
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The function of n, there is going to be n in it.
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The last thing, we evaluate the limit.
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Third, evaluate the limit, in other words, take n in that function that we get.
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This part, take n to infinity and see what you get.
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Evaluate the limit as n goes to infinity.
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That is what we are going to do, first, second, third, we will get our answer.
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We have -2 to 4, that is where we are integrating in.
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A is equal to -2.
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We know that b is equal to 4.
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We know that Δx = b - a/ n is equal to 4 - -2/ n is equal to 6/n.
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That is our Δx.
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That takes care of our Δx part, that is going to be the 6/n.
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Let us see if we can come up with something for x sub i first, some expression for x sub i,
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that we can put in to f(x), in order to get f(x sub i).
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We are going to take that, multiply it by the 6/ n to get this thing.
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That is all we are doing here.
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We have taken care of the Δx, let us see if we can find x sub i.
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Let us see if we can elucidate a pattern.
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We have x₀, that is equal to -2.
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x₁ is equal to -2 + Δx which is 6/n.
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x₂ = -2 + 6/ Δ n + another Δx which is + 12/n.
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x₃ = -2 + 12/n + another Δx which is 6/n, which is 18/n.
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Notice, this number is 1 × 6, 1, 1.
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This number is 2 × 6, 2 is here, 2 is here.
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This number is 3 × 6, 3 is here, 3 is here.
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We have our pattern, our x sub i is equal to -2 + i × 6/n.
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Which I’m going to go ahead and write it as 6i/ n.
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Great, now we found our x sub i.
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Perfect, we found our x sub i, now we are going to stick our x sub i.
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This -2 + 6i/ n into this function, to find f(x sub i).
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F(x sub i) is equal to 2 + 2 × x sub i which is equal to 2 + 2 × -2 + 6i/ n,
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that equals 2 - 4 + 12i/ n = -2 + 12i/ n, that is our f(x sub i).
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F(x sub i) × Δx is going to be -2 + 12 sub i/ n, that is my f(x sub i).
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My Δx was 6/n.
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Therefore, this is going to equal -12/n + 72/ n².
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This is my f(x sub i) Δx.
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I’m going to evaluate the sum and I'm going to take the sum as i goes from 1 to n.
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We have taken care of the first part, we are going to evaluate the sum of this.
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The sum of the f(x sub i), -12/n + 72i/ n².
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The summation symbol distributes over that.
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I get the sum i from -1 to n -12/ n + the sum of i1/ n 72i/ n².
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i is the index, i has to stay under the summation symbol.
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Everything else can be taken out as a constant.
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In other words, there is no i here.
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Therefore, I can pull out the -12n.
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Here I have to leave the i but I can pull out the 72/ n².
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Therefore, this is going to equal -12/n × the sum of 1 to n of 1 + 72/ n² × the sum i = 1 to n of i.
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Let us rewrite that so we have it on the page.
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-12/n × the sum from 1 to n of 1 + 72/ n² × the sum i from 1 to n of i.
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The sum from 1 to n of 1, we just add 1 n ×, it is just n + 72/ n².
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We have a closed form expression for this, remember.
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It is equal to n × n + 1/ 2.
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It equals -12 + 72 divided by 236, 36 × n²/ n².
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This is going to be n² + n.
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72n²/ 2n², gives me 36n² + 36n/ n².
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I just went ahead and divided the 2 into it , and then multiplied.
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72 divided by 2 is 36, 36n² 36n/ n² - 12.
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36n²/ n² is 36.
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36n/ n² is 36/n which is equal to 24 + 36/n.
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The sum of i = 1 to n of f(x sub i) dx is equal to 24 + 36/n.
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Now we evaluated the sum, we have our function of n.
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Now we take the limit as n goes to infinity.
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The limit as n goes to infinity of 24 + 36/n, this one goes to 0.
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As n goes to infinity, we are left with 24, that is our answer.
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The integral from 1 to 6 of that function, going through the entire process.
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Finding the Δx, finding the x sub i, forming f(x sub i).
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Multiplying f(x sub i) × Δx, evaluating the sum, and then taking the limit, gives us a final answer of 24.
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That is it, run through the process.
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Tedious but reasonably straightforward, as long as you have the formulas that you need.
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Evaluate the following integral using the definition, same thing.
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We know that this is going to equal, let us go ahead and write the definition,
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so that we know what we are dealing with here.
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The definition, the integral from a to b of f(x) dx equals the limit as n goes to infinity.
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I think it is a good idea to write down the definition of the equation over and over again, that way you remember it.
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The limit to infinity, the sum as i go from 1 to n of f(x sub i) Δx.
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We are going to run through the same process.
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We are going to find Δx, we are going to find an expression for x sub i.
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We are going to put x sub i into f.
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We are going to form f(x sub i).
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We are going to multiply by Δx.
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We are going to evaluate the sum of that thing.
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We are going to get a function of n and then we are going to take n into infinity to get our final answer.
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That is what we do.
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Let us go ahead and do a is equal to 1, b is equal to 6.
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Therefore, the Δx = b - a/ n which is 6 - 1/ n which is 5/n.
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That takes care of the Δx.
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x₀ that is equal to 1, x₁ = 1 + Δx which is 5/n.
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x₂ = 1 + 10/ n, x₃ = 1 + 15/n.
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We see the pattern, x sub i = 1 + 5i/ n.
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Therefore, our f(x sub i) is equal to, you put this into here, into there, f(x sub i).
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Therefore, it is 1 + 5i/ n² + 4 × 1 + 5i/ n – 7.
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It is going to equal 1 + 10i/ n + 25i²/ n² + 4 + 20i/ n - 7
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which is going to equal -2 + 30i/ n + 25i²/ n².
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This is just our f(x sub i), you need to now multiply that by our Δx which is 5/n.
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Our f(x sub i) × Δx is equal to -2 + 30i/ n + 25i²/ n² × 5/n.
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This is going to equal -10/n + 150i/ n² + 125i²/ n³.
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We have our expression, now we evaluate the sum.
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The sum from 1 to n of the f(x sub i) × Δx = the sum of 1 to n of this expression.
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-10/n + 150i/ n² + 125i²/ n³.
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Distribute, separate, sum as i goes from 1 to n of -10/ n + the sum i from 1 to n of 150i/ n²
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+ sum i from 1 to n of 125i²/ n³.
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There is no i here, pull it all out.
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I pull out the 150/ n², I pull out the 125/ n³ = -10/n × the sum i goes from 1 to n of 1 + 150/ n² ×
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the sum as i goes from 1 to n of i + 125/ n³ × the sum as i goes from 1 to n of i².
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This is going to equal -10/n × n + 150/ n² × n × n + 1/2, because that is this.
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We also have an expression for this one.
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This is going to be + 125/ n³ × n × n + 1 × 2n + 1/ 6.
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It is just arithmetic, that is all it is.
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We get -10 + 75n² + 75n/ n² + 125/ n³
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× 2n³ + 3n² + n/ 6 = -10 + 75 + 75/n + 250/6.
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I multiply everything, cancel everything.
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+ 375/ 6n + 125/ 6n² = 640/6 + 825/ 6n + 125/ 6n².
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This is our function of n, now we evaluate the limit, we take n to infinity.
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The limit as n goes to infinity of this thing, 640/6 + 825/ 6n + 125/ 6n².
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As n goes to infinity, this goes to 0, this goes to 0.
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We are left with 640/6 or 106.67.
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One painful process, we definitely need a quicker way to do this.
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We have a quicker way to do this.
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We actually done it with antiderivatives, we will do some more.
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Evaluate the following definite integral by using areas.
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6 to 30, 1/3x – 4, this is a line.
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Let us go ahead and draw this out.
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1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7,
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8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.
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Actually, I meant this to be 20 not 30.
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Therefore, I’m going to go ahead and change this to 20.
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1/3x – 4, 4 up 1/3, up 1/3, that takes us to 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
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We are going to get that line.
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We are integrating to 20.
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We are looking for the integral of this function.
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They want us to do it in terms of the areas.
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Therefore, I’m going to find area number 1 and I’m going to find area 2.
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I'm going to add them.
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Area number 1 is going to be negative because it is below the x axis.
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Area number 2 is going to be positive.
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When I add them up, I’m going to get some net area that is equal to the integral.
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Great, nice and simple.
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Therefore, this integral, the integral from 6 to 20,
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What is going on here?
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I’m not integrating from 6, I got all my numbers mixed up here.
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That is okay, very easy to fix, I’m sorry.
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The integral here is actually not from 6 to 30, it is from 0 to 20.
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I do not know where the 6 to 30 came from.
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Fortunately, everything still stands, let us recap.
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We have a function, we graph that function.
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We are integrating from 0 to 20.
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We graphed it from 0 to 20.
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Yes, we are finding this area which is going to be negative.
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We are going to be adding it to this area.
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Everything else is just fine.
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This is going to be 1/3x – 4 dx is going to be area number 1 + area number 2.
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This is 12, from 0 to 12, f(x) is below the x axis.
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Therefore, area 1 is going to be negative.
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Area 1 is going to be negative, base × height/ 2, it is just a triangle.
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Base is 12, height is 4/2.
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That gives me -24.
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Area 2 is going to be from 12 to 20, that is just going to be positive.
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It is going to be base × height/ 2.
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The base of this triangle is 8, the height is 20.
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I just put it into here and I end up with a height of 2.667/2.
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I get 10.668.
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Therefore, I just add these two together.
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a1 + a2, I get a -13.32.
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The integral of the function is a negative number, the net area.
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More area here than there is here.
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If the integral from 7 to 14 of f(x) = 27 and if the interval from 10 to 14 = 9, find the integral from 7 to 10.
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We have a nice property that takes care of this for us.
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We want the integral from 7 to 10 of f(x) d(x).
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That equals the integral from 7 to 14 of f(x) d(x) + the integral, as long as this and this are the same, the interval from 14 to 10.
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This and this are that and that.
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If these are the same, I can just add these integrals to give me this final one.
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Here we are fine, the integral from 7 to 14 of f(x) dx.
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Here, this is the integral from 14 to 10, what they gave us is 10 to 14, lower to upper 10 to 14.
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Now it is 14 to 10, it is switched.
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What I have to do is, when I switch the limits, I change the sign of the integral.
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It is - this is equal to negative of the 10 to 14 of the f(x) dx.
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Therefore, it is just equal to 27 - 9 = 18.
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