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Consider again the problem of finding the area

∫abf(x)dx

We've seen how we can do this with the trapezium rule, and Riemann sums are another way. In some ways, a Riemann sum is just a much simpler cousin of the trapezium rule. Instead of trapezia, we use rectangles:

Nevertheless, if we keep adding more and more rectangles, the Riemann sum converges to the true area just as surely as the trapeziums would. Slide n in the image above to see what happens as n (the number of rectangles) becomes large.

Let's take a closer look at a rectangle

We see its area is

f(xi−1)(xi−xi−1)

We often write Δx=xi−xi−1 for ease. When we add all the rectangles together, we get an approximation for the area:

∫abf(x)dx≈∑i=1nf(xi−1)Δx

As we add more and more rectangles, this approximation becomes more and more accurate. In fact

∫abf(x)dx=limn→∞∑i=1nf(xi−1)Δx

So this explains the notation: ∫ is the area after the ∑ of rectangles has been smoothed out, and since the rectangles become very thin, the large Δx becomes the small dx


Let

f(x)=2x2

Given the below table of values

xif(xi)−550−318−121228

calculate a 4-rectangle Riemann sum estimate for

∫−522x2dx