Current objective
Approximate the area under a curve using rectangles
Question
The graph of a function is shown below as a blue curve. Create a visualization of a left-endpoint approximation for the area under the curve on the interval [1,7] using 8 rectangles.
Slide the orange points horizontally to adjust the endpoints of the interval. Use the vertical slider on the right side of the graphing window (blue movable point) to control how many rectangles your approximation for the area will have. The value of each rectangle's width, Δx, is also shown. Finally, drag the black movable points to adjust the height of each of the rectangular boxes.
Well done! You got it right.
Answer Explanation
For this problem, we need to recall how to approximate the area under a curve by using the left-endpoint approximation method.
First, we introduce some terminology that will be useful to us. A set of points P={xi} for i=0,1,2,…,n with a=x0<x1<x2<⋯<xn=b, which divides the interval [a,b] into subintervals of the form [x0,x1], [x1,x2], …, [xn−1,xn], is called a partition of [a,b]. If the subintervals all have the same width, the set of points forms a regular partition of the interval [a,b].
We can use this regular partition as the basis of the left-endpoint approximation method for estimating the area under the curve. On each subinterval [xi−1,xi] (for i=1,2,…,n), we construct a rectangle with width Δx and height equal to f(xi−1), which is the function value at the left endpoint of the subinterval. Then, the area of this rectangle is f(xi−1)Δx. Adding the areas of all these rectangles, we get an approximate value for the area, A, under the curve in the interval [a,b]. We use the notation Ln to denote that this is the left-endpoint approximation of A using n subintervals. That is, A≈Ln=f(x0)Δx+f(x1)Δx+⋯+f(xn−1)Δx=n∑i=1f(xi−1)Δx.
We must first select the correct endpoints (orange movable points) before approximating the area under the curve. The interval is given as [1,7], therefore the left endpoint is a=1 and the right endpoint is b=7. Next, we must adjust the number of rectangles to use for estimating the area under the curve. We are asked to use n=8 rectangles, so we slide the blue point on the right side of the graph window vertically until the correct number of rectangles is shown. Since the width of the entire interval is b−a=7−1=6 and we will use 8 rectangles to approximate the area, the width of each rectangle is Δx=b−an=68=0.750.
We must now adjust the height of each of the rectangles so that we obtain the left-endpoint approximation of the area under the curve. This means that the height of each rectangle should correspond to the value of the function at the left endpoint of the rectangle. Furthermore, since the interval [1,7] has been divided into subintervals of equal width (a regular partition of the interval), successive left-endpoints of the rectangles can be found by adding multiples of Δx=0.750 to a=1. Specifically, using the notation we introduced above for the sequential values of x, the x-coordinates of the left endpoints are x0=a=1x1=a+Δx=1.75x2=a+2Δx=2.5⋮x7=a+7Δx=6.25.
The area, A, can then be written as follows: A≈L8=f(x0)Δx+f(x1)Δx+f(x2)Δx+f(x3)Δx+f(x4)Δx+f(x5)Δx+f(x6)Δx+f(x7)Δx=f(a)Δx+f(a+Δx)Δx+f(a+2Δx)Δx+f(a+3Δx)Δx+f(a+4Δx)Δx+f(a+5Δx)Δx+f(a+6Δx)Δx+f(a+7Δx)Δx.
Note that this sum can be written using summation notation as A≈L8=8∑i=1f(a+(i−1)Δx)Δx.
Substituting the values for x that represent the left endpoints of the rectangles and factoring out Δx=0.750, we get A≈L8=f(1)0.750+f(1.75)0.750+f(2.5)0.750+f(3.25)0.750+f(4)0.750+f(4.75)0.750+f(5.5)0.750+f(6.25)0.750=(f(1)+f(1.75)+f(2.5)+f(3.25)+f(4)+f(4.75)+f(5.5)+f(6.25))0.750.
Here, we are not asked to compute this approximation for the area because we were not given the precise mathematical form of the function f(x). We have, however, shown how this area can be calculated and constructed a visualization for this approximation on the interval [1,7].