In the previous activity we saw that as the number of samples increased, not only did our estimate of the car count become more accurate, but the estimate also converged towards the area under the function [code]g(x)[/code]. The intuitive conclusion we can draw is that the area under the function [code]g(x)[/code] is the result of taking "infinitely many" samples, and is therefore the best estimate of car count between 6am and 4pm that we can obtain from the model. This observation gives significance to the area under the function [code]g(x)[/code]. Indeed, [i]this[/i] concept of the area under a curve [i]is[/i] the integral.[br][br][b]Definition[/b]: The area between the graph of a function [code]f(x)[/code] and the x-axis, and between the vertical lines [code]x=a[/code] and [code]x=b[/code] is called the [b]integral[/b] of [code]f(x)[/code] between [code]x=a[/code] and [code]x=b[/code]. [b]Caveat[/b]: Any area below the x-axis will be counted as negative.[br][br]The mathematical shorthand for the integral of [code]f(x)[/code] between [code]x=a[/code] and [code]x=b[/code] is[br][br][math]\int^{x=b}_{x=a}f\left(x\right)dx[/math][br][br]Or sometimes the "x="s are omitted, and integrals are written as[br][br][math]\int^b_af\left(x\right)dx[/math][br][br]DO NOT concern yourself with the mysterious "dx". We'll discuss that later. For now, your best bet is to think of the above expression as shorthand for "the integral of [code]f(x)[/code] from [code]x=a[/code] to [code]x=b[/code]." The "dx" is silent, if you will.[br][br]The applet below helps you visualize the definition of the integral for the function [code]f(x)=x^3-6x^2+9x+2[/code], which we saw [url=https://www.geogebra.org/m/x39ys4d7#material/mznjt3yx]earlier in a lesson[/url] about maximizing and minimizing functions. You can click and drag [code]a[/code] and [code]b[/code]. As you change [code]a[/code] and [code]b[/code], you'll notice the value of the integral changes. At the outset the integral is 8, but adjustments to [code]a[/code] and [code]b[/code] will of course impact area of the shaded region, and therefore will also impact the value integral. For instance when [code]a[/code] and [code]b[/code] are close together, the integral will be smaller because the shaded region is smaller. Another thing to try is sliding [code]a[/code] to -2 and [code]b[/code] to -1; the shaded region is now entirely below the x-axis, and so the integral becomes negative. This is the "caveat" in the definition above. We'll discuss these nuances and others in the next lesson.
Quick Check: Use the applet to calculate[br][br][math]\int_2^4x^3-6x^2+9x+2dx[/math]
Before we get into nuances, let's take a look back at g(x), the model of the rate of traffic along Route 15 in Johnson Vermont. This applet helps you visualize [i]both[/i] the integral of the function [code]g(x)[/code] (maroon) and rectangle estimate (green) with [code]Samples[/code] set to 5. [br][br]On the left is the sum of the areas of the rectangles which is only a rough estimate of the car count on Route 15. On the right is the integral of [code][/code][code]g(x)[/code] from [code]x=StartTime [/code]to [code]x=FinishTime[/code] which is model's best estimate of the car count on Route 15. As noted earlier, you can think of the integral of [code]g(x)[/code] as if we took an infinite number of samples of [code]g(x)[/code] between [code]x=StartTime[/code] and [code]x=FinishTime[/code] .[br][br]You can now also adjust [code]StartTime[/code] and [code]FinishTime[/code] to explore the consequences. For instance, if you put the two close together, the car count (which [i]is[/i] the integral) will of course decrease because fewer cars roll along in a shorter period of time. [br][br]So long as [code]StartTime[/code] is before [code]FinishTime[/code], this integral will always be positive because g(x) is always positive. This also makes sense because the rate of traffic on a highway can only ever be positive. Don't concern yourself with the oddity of if [code]FinishTime[/code] is before [code]StartTime[/code] just yet. That's for later. Keep it simple for now.
Use the applet above to calculate[br][br][math]\int_{400}^{1000}5.96076+5.10899\sin\left(0.00429x-2.0721\right)dx[/math]
Move forward to the next activity where we'll discuss a few of the finer points about integrals and the rectangle approximations we've been encountering.