WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to be doing some example problems for areas between curves.
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Let us jump right on in.
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In the following examples, we want you to sketch the given curves, identify the region that is enclosed by these curves.
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State whether you are going to be integrating with respect to x or y.
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Draw a representative rectangle, if you can.
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Finally, find the area of this region, or at the very least set up the integral for the area of this region.
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Let us get started.
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Our first one is going to be y = 7x - x² and y = x.
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Let us go ahead and draw this out and see what is it we are actually dealing with.
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I think I will go ahead and work in blue.
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Let me draw it over here.
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How we are going to draw 7x - x²?
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I'm going to go ahead and work over here.
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I’m going to factor this, y = x × 7 - x and that is equal to 0, that gives me x = 0 and that gives me x = 7.
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I know that the graph is going to hit a 0 and I know the graph is going to hit at 7.
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The midpoint between 0 and 7, the midpoint which is where the vertex is going to be,
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I know this is a -x², I know it is a thing that is going to open down.
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Therefore, I know it is going to be going up this way and up this way.
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The midpoint = 3.5.
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When I put 3.5 into the original, I end up with y = 12.25.
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At 3.5, 12.25, that is that graph right there.
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As far as the y = x is concerned, we know that one, that is just a line that way.
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The region that we are interested in is this region right here.
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In this particular case, I’m going to go ahead and draw a representative rectangle.
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The rectangle is vertical, I'm going to be adding the rectangles this way.
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I'm going to be integrating along x.
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What is the x value of this point?
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We are going to be integrating from 0 all the way to the x value of this point, whatever that is.
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We need to find that.
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What is the x value of this point?
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I’m going to set them equal to each other.
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I have got 7x - x² is equal to x.
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I have got x² - 6x = 0.
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I have got x × x - 6 = 0 which gives me x = 0.
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That is one of them.
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x = 6, that is the other point.
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Remember, this one was 7.
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We integrate from 0 to 6, there we go.
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From here to here, we are going to add up all of the individual rectangles.
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That is going to give us the area of the curve.
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Let us go ahead and do that.
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With each of these, I actually rendered the picture itself.
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We have a nice drawing to look at.
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This is the region that we are interested in, in between here and here.
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This is our 0, this is our 6, therefore, the area is going to be the integral from 0 to 6, upper function - the lower function.
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The upper function is 7x - x² - the lower function which is x.
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I’m going to integrate along x so it is dx.
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This is the integral from 0 to 6 of 7x – x.
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I can simplify 6x - x² dx.
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This is going to equal 6x²/ 2 – x³/ 3.
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I'm going to evaluate this from 0 to 6.
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My answer is going to be 36, that is it.
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I put 6 into here.
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I put 6 into here and I end up with 108 – 72, that is for the 6.
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And then, -, 0 – 0 is for the 0, that gives me my 36.
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Again, you can go ahead and actually evaluate it.
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You can go ahead and use your calculator.
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Remember, we looked at that earlier.
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Just enter the function, do second calc.
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Go down to integration, lower limit, upper limit.
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You have got yourself your integral value.
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In this particular case, it is really easy to evaluate without the calculator, but in case you want to do that.
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You got 36.
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Example number 2, we have x = y² – 3, x = e¹/2 y.
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We have y = -1 and y = 2.
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In this particular case, it is y that is the independent variable.
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We are going to be looking at functions which are left to right, instead of up down.
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Let us take a look at what it is that we have got going here.
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x = y² – 3.
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My y² graph looks like this, the -3 part means it is a shift to the left because it is a function of y.
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It is with left, my graph is actually going to be 1, 2, 3.
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It is going to look something like that.
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x = e ^½ y, when y is equal to 0, e⁰ is 1, that means x is 1.
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It is going to be one of my points.
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Let me go ahead and mark 1, 2.
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Let us go 1, 2, let us go 3.
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When y = 2 it is 2/2, it is 1 e ⁺12.718.
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When y = 2, I got to x 0.718, it is going to be somewhere around there.
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Basically, what I’m going to get is this, it is not going to cross, it is asymptotic right there.
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It is going to look something like that.
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y = -1 that this line, y = 2 is this line.
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The region that I'm interested in is going to be this region right here.
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I have a better picture on the next page.
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It is going to be something like that.
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We are going to be integrating from -1 to 2.
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We are going to be integrating along the y axis.
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Our representative rectangle is going to look something like that.
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I’m going to be integrating vertically.
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We will say best to integrate along the y axis from -1 to 2.
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Therefore, the area is going to be the integral from -1 to 2, right function - left function.
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We are integrating along y.
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Let us take a look at a better picture here.
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Yes, this is the region we are interested in.
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That, that, that, there.
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We want this and we are integrating horizontally from -1 to 2.
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Therefore, our area is going to equal the integral from -1 to 2.
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The right function is going to be e ^½ y - the left function y² - 3 dy.
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That is going to equal the integral from -1 to 2 of e ^½ y - y² + 3 dy.
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That is going to equal 2 e ^½ y - y³/ 3 + 3y evaluated from -1 to 2.
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I went ahead and I just did this via my calculator.
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However, if you want to see what it actually looks like, when you plug 2 in here, you are going to get 3 terms.
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-1 in here, you are going to get 3 terms.
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I got 2e – 8/3 + 6, that is when I plug the 2 in, and then I subtract.
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When I plug in the -1, it is going to be 2e⁻¹/2 + 1/3 – 3.
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If I have done everything correctly, and my final answer that I got was 10.224.
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That is it, nice and straightforward.
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Right function - left function, in this case.
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y = 1/x, y = 1/x³, x = 4.
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This one should be reasonably straightforward, let us do this
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Let us go 1, 1, 1/x.
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Let me make this a little bit bigger.
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I’m going to put the 1,1 right there.
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I have got something like that, that is my y = 1/x, my 1/x³.
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Something like that, that was my y = 1/x³.
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If I go to 4, something like that.
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It looks like I’m going to be integrating from 1 to 4.
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I'm going to be integrating, this is one of my representative rectangles.
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I'm going to be adding the rectangles horizontally, I’m integrating along the x axis.
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We integrate along x from 1 to 4.
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Let us take a look at a picture here.
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Nice, better, looking picture.
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This is the region, here is our 1, here is our 4.
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Here is our representative rectangle and we are adding this way.
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What I have got is the area = the integral from 1 of 4 of the upper function - the lower function.
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The upper function is this one, that is the 1/x - 1/x³ dx.
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It is going to be integral from 1 to 4 of 1/x dx - the integral from 1 to 4 of x⁻³ dx.
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It is going to equal, this one is going to be ln(x) evaluated from 1 to 4, -x⁻²/ -2 evaluated from 1 to 4.
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If I have done everything correctly, I get ln 4 - ln 1 - -1/32 - -1/2, this goes to 0.
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When I add everything up, I get 0.9175.
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Again, it all comes down to the same thing.
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It is going to be the upper function – the lower function or right function - left function.
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All you have to do is find the limits of integration.
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The limits of integration are going to come from either ends that are given to you, in this case x = 4.
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The x value or the y value for where the two graphs happens to meet.
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That is the only thing that is going on.
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15 – 2x², y = x² – 5, I mentioned the biggest difficulty here is actually drawing these functions out,
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if you remember them from pre-calculus.
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Let us go ahead and draw this out.
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15 - 2x² - 2x² opens down, let us see where it actually hits the x axis.
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We have got 15 - 2x² is equal to 0.
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You have got 2x² = 15, x² = 15/2, that means x is going to be + or -2.74.
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It is going to be, I have got a point here, a point here.
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I have one parabola that looks like this.
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I have my other parabola which is going to be the x² – 5.
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x² - 5 = 0, x² = 5, x = + or -2.24.
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2.24 is like there and there, down to 5.
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This parabola goes this way, this way.
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I need the area in between those two curves.
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It is going to be upper – lower.
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Here is going to be one of my representative rectangles.
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I'm going to integrate along the x axis.
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My limits of integration, I need to know the x value of that point and the x value of that point, where those two meet.
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Let us find that out, I find that out by setting the two functions equal to each other.
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15 - 2x² = x² – 5.
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I get 20 = 3x², x² = 20/3, x = + or -2.582.
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This is 2.582, this is -2.582.
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Those are going to be my limits of integration.
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I'm going to take the integral, the area is going to be integral from - 2.582 to +2.582.
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The upper function - the lower function dx.
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Let us see what that looks like.
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There you go, here is your -2.582, here is your +2.582.
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I have got my area is equal to the integral -2.582.
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You are definitely going to need to use a calculator for this.
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2.582 of the upper function which is 15 - 2x² - the lower function which is x² - 5 dx,
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which is equal to the integral -2.582 to +2.582.
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It looks like I have got 20 - 3x² dx.
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This is going to equal 20x - x² evaluated from that to that.
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When I put the values in, I get 68.853.
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There you go, that is the area between the curves.
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Nice, simple, straightforward, not a problem.
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x = 1/8 y³, x = 6 - y², let us go ahead and draw this out.
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Again, it is probably best to just go ahead and use your calculator or something like www.desmos.com.
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What you end up getting is, 6 - y² this is going to be a graph that looks like this.
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x = ½ y³ is going to look something like this, something like that.
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We are looking at this region right there.
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It looks like it is best to go horizontal rectangles, which means we are going to integrate from bottom to top.
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We want to integrate along y, we need the y value of that point and we need the y value of that point.
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That is going to be our lower limit to our upper limit of integration.
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Let us go ahead and do that, find that first.
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We set the two functions equal to each other.
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1/8 y³ = 6 - y².
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I get y³ + 8y² - 48 is equal to 0.
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We use our calculators, we use our software, we use Newton’s method,
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whatever it is that we need to do to find the values of y to satisfy that.
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We end up with two values, y = 2.172 and y = -3.144.
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This value is our 2.172, that is our upper limit of integration.
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This value is our -3.144, that is our lower limit of integration.
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Our area is going to equal -3.144 to 2.172, right function - left function.
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We are integrating along y, this is going to be, dy.
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Here is the region that we are discussing.
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In between there, that is our y value of 2.172.
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Here is our y value of -3.144.
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What have we got?
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Let me go ahead and write it down here.
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We have the area is equal to the integral -3.144 to 2.172, right function - left function.
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I have got 6 - y² – 1/8 y³ dy.
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I get 6y - y³/ 3 – y⁴/ 32.
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I evaluate this from -3.144 to 2.172.
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When I put it into a calculator, I get 20.479.
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Nice and straightforward.
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Let us do something that involves some sine and cosine.
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We have y = cos x, y = sin 2x.
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I would like you to integrate this or find the area the region between those two curves, between 0 and π/2.
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Let us go ahead and draw this out.
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Let us stay in the first quadrant here.
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That is okay, I do not need to make it quite so big.
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y = cos x, period is 2π, it is going to look something like this.
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This is π/2.
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y = sin(2x), the period is π.
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If the period is π, that means it is going to start at 0 and
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it is going to hit 0 again at π/2 that means at π/4, it is going to hit a high point.
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Let us go ahead and call this 1.
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We are going to get something like that, the region that we are interested in, this region right here.
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What is the area of that region?
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Notice, here this function, this is our cos x.
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This function is our sin 2x, from 0 to some value which I will call a for now, cos x is above sin x, that is upper – lower.
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Pass this point, it is the sin 2x that is actually the upper function and the cos x is the lower function.
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I'm going to have to break this up into two integrals.
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Integrate from 0 to a, cos x - sin 2x, from a to π/2, sin 2x - cos x.
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We need to find what a is first.
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How do you find what a is equal to?
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You set the two functions equal to each other and you solve for x.
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I have got cos(x) = sin(x), cos(x) -, sin(2x) there is an identity.
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Sin(2x) = 2 sin x cos - 2 sin x cos x = 0.
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I have moved it to the left and used my identity.
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I’m going to factor out a cos x.
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Cos x, 1 – 2, sin x = 0 that gives me two equations.
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Cos x = 0 and sin x = ½.
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Cos x = 0 that is going to be my π/2.
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I already know that they meet there.
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This value right here, sin(x) = ½.
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Between 0 and π/2, my x = π/ 6, that is equal to my a.
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My a is equal to π/6.
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Therefore, I’m going to be integrating from 0 to a, 0 to π/6.
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I’m going to be integrating from π/6 to π/2.
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This first integral, second integral.
00:28:04.100 --> 00:28:07.900
Let us go ahead and see what this looks like.
00:28:07.900 --> 00:28:19.200
The region that I’m interested in is, a that is going to be my first integral.
00:28:19.200 --> 00:28:23.000
This is going to be my second integral.
00:28:23.000 --> 00:28:31.200
The red one is my cos(x), the blue one is my sin(2x).
00:28:31.200 --> 00:28:46.600
Therefore, what I have got is area is equal to the integral from 0 to π/6 cos x - sin(2x) dx
00:28:46.600 --> 00:28:51.700
+ the integral from π/6 to π/2.
00:28:51.700 --> 00:29:00.800
Now it is sin 2x - cos x.
00:29:00.800 --> 00:29:03.200
Nice and straightforward, the rest is just integration.
00:29:03.200 --> 00:29:20.900
This is going to be sin x + ½ cos 2x evaluated from 0 to π/6 +
00:29:20.900 --> 00:29:35.700
this is going to be -1/2 cos(2x) - sin x evaluated from π/6 to π/2.
00:29:35.700 --> 00:29:41.200
When you actually evaluate this which I would not do here, you are going to get ½.
00:29:41.200 --> 00:29:55.100
If I have done all of my arithmetic correctly, or if my calculator do the arithmetic correctly.
00:29:55.100 --> 00:30:00.000
y = 2x² 10 x² 7x + 2y = 10.
00:30:00.000 --> 00:30:05.600
You know what, this example, I think I'm actually just going to go through really quickly,
00:30:05.600 --> 00:30:08.800
just run through it because I think we get the idea now.
00:30:08.800 --> 00:30:12.300
Do not want to spend too much time hammering the point.
00:30:12.300 --> 00:30:15.200
This is what the graph looks like.
00:30:15.200 --> 00:30:22.500
Here is your graph of the y = 10x².
00:30:22.500 --> 00:30:25.400
This is the 2x².
00:30:25.400 --> 00:30:36.400
This time right here is the 7x + 2y = 10 which I have actually written as y = -7/2 x + 5.
00:30:36.400 --> 00:30:40.200
You are going to solve it for y = mx + b, in order to graph it.
00:30:40.200 --> 00:30:49.200
The region we are interested in is this region right here.
00:30:49.200 --> 00:30:52.900
I’m going to go ahead and do this by breaking this up.
00:30:52.900 --> 00:30:54.400
I’m going to do this is in two integrals.
00:30:54.400 --> 00:30:57.600
I’m going to integrate along x.
00:30:57.600 --> 00:31:04.600
I’m going to take the representative rectangle there for my first region and my second region.
00:31:04.600 --> 00:31:09.800
I’m going to go from 0 to whatever this number is, which I will find in just a minute.
00:31:09.800 --> 00:31:18.800
I'm going to go from this number to whatever this number is, for my second integral.
00:31:18.800 --> 00:31:33.100
When I set y = 10x² equal to y = -7/2 x + 5, I’m going to end up this value right here.
00:31:33.100 --> 00:31:38.200
My x value is going to be 0.553.
00:31:38.200 --> 00:31:44.700
This point, I'm going to set my 2x² equal to my -7/2 x + 5.
00:31:44.700 --> 00:31:57.400
This value I'm going to get is going to be 0.932.
00:31:57.400 --> 00:32:00.300
Once I have that, the integral is really simple.
00:32:00.300 --> 00:32:14.100
It is just going to be the area is equal to the integral from 0 to 0.553, upper function - lower function,
00:32:14.100 --> 00:32:35.500
10x² - 2x² dx + 0.553 to 0.932 of the upper function which is -7/2 x + 5.
00:32:35.500 --> 00:32:42.800
Because we are integrating with respect to x, it has to be a function of x - the lower function.
00:32:42.800 --> 00:32:48.100
From here to here, that is the lower function which is the 2x² dx.
00:32:48.100 --> 00:32:59.700
When I evaluate this integral and solve, I get 0.9341 as my area, that is it.
00:32:59.700 --> 00:33:02.800
This is region 1, this is region 2, I just broke it up.
00:33:02.800 --> 00:33:09.800
I have to find the x value for there, the x value for there, integrate from 0 to that point first
00:33:09.800 --> 00:33:13.600
and that point first, upper – lower, upper – lower.
00:33:13.600 --> 00:33:20.100
That is it, just do what is exactly what you think you should do.
00:33:20.100 --> 00:33:25.900
It is very intuitive.
00:33:25.900 --> 00:33:28.000
Let us go ahead and talk about this problem now.
00:33:28.000 --> 00:33:32.600
Cars A and B start from rest and they accelerate.
00:33:32.600 --> 00:33:37.900
The graphs below show their respective velocity vs. time graph × along the x axis,
00:33:37.900 --> 00:33:41.700
the velocity of the cars is along the y axis.
00:33:41.700 --> 00:33:45.200
Car A is the blue graph, this is A.
00:33:45.200 --> 00:33:55.600
The graph, the value of the function is, this function right here is √2x.
00:33:55.600 --> 00:33:59.100
B is the red graph, this is car B.
00:33:59.100 --> 00:34:07.500
Its function, when expressed as a function of x is 1/5 x³.
00:34:07.500 --> 00:34:12.500
First question we are going to ask, at 2.187 minutes where the graphs meet,
00:34:12.500 --> 00:34:21.400
this point right here, this is our 2.187, which car is further ahead?
00:34:21.400 --> 00:34:27.100
This is a velocity vs. time graph, when you integrate a velocity vs. time graph,
00:34:27.100 --> 00:34:34.900
in other words, when you find the area under the curve up to a certain time, that gives you the total distance traveled.
00:34:34.900 --> 00:34:47.700
Because again, velocity is, let us say meters per second, the differential time element is dt, meters per second.
00:34:47.700 --> 00:34:56.100
The integral of v dt, velocity is expressed in, let us just say it is meters per second.
00:34:56.100 --> 00:34:58.700
Time is expressed in second.
00:34:58.700 --> 00:35:04.800
When you multiply those and add them all up, in other words integrate from 0 to 2.187,
00:35:04.800 --> 00:35:08.600
you are going to get the area under the curve for car A.
00:35:08.600 --> 00:35:16.700
You are going to get the area under the curve for car B, whichever area is bigger, that has gone further.
00:35:16.700 --> 00:35:27.200
Clearly, car A has the bigger area, the area under all of the blue graph is a lot more than the area under the red graph.
00:35:27.200 --> 00:35:34.400
Therefore, part A is really simple, it is car A is further ahead.
00:35:34.400 --> 00:35:39.900
Car A is further ahead and this is further ahead because
00:35:39.900 --> 00:35:55.300
the area under its graph is a lot more than area of the graph for car B which is just that right there.
00:35:55.300 --> 00:36:02.200
If we shaded the region between the graphs from t = 0 to t = 187, what would the shaded area represent?
00:36:02.200 --> 00:36:03.900
Let me go to black.
00:36:03.900 --> 00:36:15.000
If I shaded in the area between the graphs, in other words this area, if I shaded that area, what does that represent?
00:36:15.000 --> 00:36:19.300
The area on the blue graph is the total distance the blue has gone.
00:36:19.300 --> 00:36:25.100
The area under the red graph is the total distance the red car has gone.
00:36:25.100 --> 00:36:33.600
The difference between them is just how much further car A is then car B.
00:36:33.600 --> 00:36:38.600
Part B, I will just write it up here.
00:36:38.600 --> 00:36:43.800
Part B is what does it represent?
00:36:43.800 --> 00:36:58.400
The shaded region represents the distance A is farther from the farther from B.
00:36:58.400 --> 00:37:05.600
That is all, area under the graph for A, area under the graph for B,
00:37:05.600 --> 00:37:07.900
the difference between them is the area between the two graphs.
00:37:07.900 --> 00:37:12.900
It is how much father A is than B.
00:37:12.900 --> 00:37:21.300
Part C, at 4 minutes, which car is ahead, and D, at what time will the cars be side by side?
00:37:21.300 --> 00:37:28.600
When we look at this graph, we see that A, it accelerates faster and steadies out.
00:37:28.600 --> 00:37:34.900
B, accelerates slower, it is further behind but at some point it really starts to accelerate.
00:37:34.900 --> 00:37:36.600
Eventually, it is going to catch up.
00:37:36.600 --> 00:37:41.600
A is going to be ahead of be but at some point B, is going to pass A.
00:37:41.600 --> 00:37:43.900
At 4 minutes, which car is ahead?
00:37:43.900 --> 00:37:52.000
The question we are asking is at 4 minutes which has a greater area under its graph? 4.
00:37:52.000 --> 00:37:57.000
Let us go ahead and work that one out on the next page with a graph.
00:37:57.000 --> 00:38:03.100
We are going to integrate, we are going to find the area under the blue graph from 0 to 4,
00:38:03.100 --> 00:38:07.800
and we are going to find the area under the red graph from 0 to 4.
00:38:07.800 --> 00:38:18.800
We said that the blue graph was √2x and we said this one was 1/5 x³.
00:38:18.800 --> 00:38:39.100
From car A which is the blue graph, we have the area is equal to the integral from 0 to 4 of √2x dx.
00:38:39.100 --> 00:38:44.600
When I solve that, I get 7.543.
00:38:44.600 --> 00:38:56.000
For car B, the area = the integral from 0 to 4 of 1/5 x³ dx.
00:38:56.000 --> 00:39:00.600
When I solve that, I get 12.8.
00:39:00.600 --> 00:39:07.600
Car B, at 4 minutes or 4 seconds, whatever the time unit is, now car B has actually past car A.
00:39:07.600 --> 00:39:15.200
Car B is farther forward.
00:39:15.200 --> 00:39:21.600
The last question asked, at what time are they side by side?
00:39:21.600 --> 00:39:24.400
Side by side means it is going the same distance.
00:39:24.400 --> 00:39:28.200
At what time are the two areas equal?
00:39:28.200 --> 00:39:38.900
We need the area of car A to equal the area of car B.
00:39:38.900 --> 00:39:44.500
The area of car A is going to be the integral from 0 to l.
00:39:44.500 --> 00:39:55.200
L is the time that we are looking for of √2x dx is equal 0 to l, that is the time period.
00:39:55.200 --> 00:40:07.100
From 0 to whatever time we are looking for, we are looking for l, 1/5 x³ dx.
00:40:07.100 --> 00:40:14.900
This is going to be √2 × the integral from 0 to l of x ^½ dx.
00:40:14.900 --> 00:40:26.000
It is going to equal 1/5, the integral from 0 to l of x³ dx.
00:40:26.000 --> 00:40:50.900
This is going to be, I’m going to come up here, 2√2/ 3 x³/2 from 0 to l is going to equal x⁴/ 20 from 0 to l.
00:40:50.900 --> 00:40:57.600
When I put l in for here, when I put l in for here, 0 on for here, 0 on for here, set them equal to each other and solve.
00:40:57.600 --> 00:41:12.600
I get something like this, I get 2√2/ 3 l³/2 = l⁴/20.
00:41:12.600 --> 00:41:15.200
I need to solve for l.
00:41:15.200 --> 00:41:37.800
The equation that this gives me is l⁴/20, let us bring this over this side, set it equal to 0 – 2√2/ 3 l³/2 = 0.
00:41:37.800 --> 00:41:52.100
When I use my mathematical software or my calculator, or whatever it is to find what l is, I get l = 3.24 seconds.
00:41:52.100 --> 00:41:53.200
I hope that made sense.
00:41:53.200 --> 00:41:57.600
I want to know, when are they going to be side by side?
00:41:57.600 --> 00:41:59.900
Side by side means they have gone the same distance.
00:41:59.900 --> 00:42:03.400
Distance is the area underneath their respective graphs.
00:42:03.400 --> 00:42:08.600
I set the areas equal to each other, the areas are the integrals.
00:42:08.600 --> 00:42:11.600
L is my time, the upper limit integration.
00:42:11.600 --> 00:42:14.400
I’m solving for l, I get an equation in l.
00:42:14.400 --> 00:42:17.100
I solve for l, I get 3.24 seconds.
00:42:17.100 --> 00:42:20.200
I did this by using a particular graph.
00:42:20.200 --> 00:42:22.000
Here is what the graph looked like.
00:42:22.000 --> 00:42:36.200
That function that we ended up getting, that this function is our l⁴/20 – 2√2/ 3 l³/2 = 0.
00:42:36.200 --> 00:42:40.300
I just want to know where it is equal 0, 3.24 seconds.
00:42:40.300 --> 00:42:46.700
You can use a calculator, you can use a graph, find where it crosses the x axis, anything you want.
00:42:46.700 --> 00:42:52.200
I hope that make sense, and that is areas between curves.
00:42:52.200 --> 00:42:54.200
Thank you so much for joining us here at www.educator.com.
00:42:54.200 --> 00:42:55.000
We will see you next time, bye.