(a) Graph the function f(x) = x - 2 lnx 1 ≤x ≤5 (b) Estimate the area under the graph of f usin

(a) Graph the function f(x) = x - 2 lnx 1 ≤x ≤5 (b)...

Key Concepts

Graphing Functions

Graphing functions involves plotting the behavior of a function over a specified domain. This process requires understanding the domain restrictions, intercepts, increasing or decreasing intervals, concavity, and potential asymptotes or boundaries. In the context of calculus, properly visualizing a function helps in understanding its overall behavior, which is essential for tasks like area estimation under the curve.

Riemann Sums

Riemann sums are a fundamental concept in numerical integration used to approximate the area under a curve by dividing the region into a finite number of rectangles. Each rectangle's area is computed using a sample point from the subinterval, and the sum of these areas provides an estimate of the definite integral. This method lays the groundwork for understanding more advanced integration techniques.

Sampling Methods in Numerical Integration

Sampling methods in numerical integration refer to the selection of specific points within each subinterval (like right endpoints or midpoints) to calculate the height of each approximating rectangle. The choice of sample points affects the accuracy of the approximation. Using different sampling methods can lead to varying degrees of error and is a key consideration when estimating integrals.

Subdivision and Convergence

Increasing the number of rectangles, or subintervals, improves the approximation of the area under a curve by reducing the error inherent in the rectangular approximation. This refinement process is central to the concept of convergence in integration, where the Riemann sum approaches the exact value of the definite integral as the subintervals become infinitesimally small.

Transcript

00:03 So in this problem, we have a function, f of x is x minus 2, natural log of x.

00:08 First part was to graph it.

00:09 So whatever tools you use, if you're graphing on paper, if you've got your favorite graphing tool, we see the graph here.

00:14 And what we're asked to do is to use rectangles to approximate the area under the curve.

00:20 So the first thing in the first part is to use four rectangles.

00:24 Okay.

00:24 So if i'm going to use four rectangles, i'm trying to approximate the area that goes from one over to five.

00:31 So if i use four rectangles, each rectangle is one in length.

00:36 And they say use the right point of each interval to determine the height of that rectangle.

00:41 So what that means is you're going to have f of 2, f of 3, f of 4, and f of 5 to determine the height of the rectangles.

00:48 So the first part is going to be the width of the rectangle.

00:53 So the width is going to be one.

00:56 And then the height of each rectangle.

01:00 So the height is going to be determined by f of the rectangle.

01:02 2 plus f of 3 plus f of 4 plus f of 5 and that will give me my approximation to the area using four rectangles so let's take a look at this on our calculator so if i store the function i've stored the function f of x so f of x sorry about that so i've stored f of x as x minus 2 natural or log of x so it's one times and then that's going to be f of two plus f of three plus f of four plus f of five and that gives me four point four two five so this first estimate with four rectangles right in points four point four two five so this is about four point four four four uh two five now they ask us to keep those same for rectangles, but instead of using the right point, use the midpoint.

02:13 So the midpoints will be happening at 1 .5, 2 .5, 3 .5, 4 .5.

02:20 So it's going to be one times, and this is going to be f of 1 .5 plus f of 2 .5, plus f of 3 .5, plus f of 4 .5.

02:37 So that's using the midpoint of each rectangle.

02:40 So let's just go back and look at that on our calculator.

02:44 And what i'll come up with there is i can just recall this command.

02:50 And it's just f of 1 .5, f of 2 .5, f of 3 .5, and f of 4 .5.

03:05 So i get the approximation of 3 .84.

03:08 So 3 .84 is my approximation using midpoints.

03:16 Okay.

03:17 If i want to find out how i'm doing, i looked at this real quick.

03:20 If i looked at the actual value of this definite integral is about 3 .91 or 3 .905.

03:28 So the exact value is, so the exact value is close to what was that, 3 .905.

03:44 So 3 .905.

03:47 So what i see so far is the midpoints with four rectangles was better than the right side of each rectangle.

03:55 What they do in the next part is to say let's do this same exercise but with eight rectangles...

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