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# Motion problem with Riemann sum approximation

Here we see how estimating the area under a curve representing velocity estimates distance traveled. Created by Sal Khan.

## Want to join the conversation?

• In short, the more rectangles we come up, the better approximation we can get? But why only rectangles, not rectangles & triangles? For example: We all know `Δx = x₁ - x₀` and the area of the first rectangle is `A = f(x₁) * Δx`. How about `A = f(x₀) * Δx + 1/2 * [f(x₁) - f(x₀)] * Δx`?

P/S: If I simplify that hairy expression, it becomes `A = Δx * [1/2 * (f(x₀) + f(x₁))]`.
• Good thinking. Actually, that is one of the formulas. More specifically, the trapezoidal area formula. If you rewrite your formula, it is height*(1/2(base 1 + base 2)).
• If the graph was showing the acceleration instead of the velocity, would the area under the curve still be the distance traveled?
• No, the area under the curve would be the velocity.
• Would the average of right rectangles and left rectangles be a more accurate approximation than the two individually?
• if you take right handed and left handed rectangles then take average of those two, would you get an exact value of the total distance traveled?
• I would recommend looking at a graph while doing this. On your graph, pick a x sub 0 and a x sub 1 (as your width.) Then draw a rectangle with it's height being the average height of that graph within your interval. When we look for the exact area under a curve, we don't want our rectangle to be above or below the curve at any point. As you can see when you draw the rectangle, parts of it exceeds the curve and parts of it under exceeds the curve. Taking the average of the heights for your rectangle would give you a better approximation of the actually area under the curve, but it would still not give you the actually area under the curve.
• Wouldn't it be more accurate to do this with triangles and trapezoids? Why don't we do it like that?
• Would you get the exact distance traveled if you took the average of the over estimation plus the under estimation?
(1 vote)
• Just image a parabula y = x^2 from 0 to 1:

Overestimate with 1*1= 1 and underestimae with 0*1=0

So the average would be 1/2(0+1) = 1/2 , wich is the exact area of a triangle and not the curve under the parabula.
• If the right riemann sum is an overestimate of the area why don't we take the limit of the curve to precisely approximate the area under the curve like we did in derivatives. Please correct me if I am wrong.
(1 vote)
• In this example we can't take the limit of the curve, because we don't have a continuous function for the curve: we only know what the velocity is at two second intervals, not what happens in between. However, if you do have a continuous function, then yes, you can decrease the width of the rectangles to the limit of width = 0. That's where this tutorial is heading.
• Is there a way to caculate the exact number of the area?
• Yes. That's what the rest of the integral calculus playlist is about. The gist is that we take the limit of a sequence of better and better approximations.
• In this specific problem, could you take both a left and a right-handed limit, find the delta, subtract it from the left-handed limit, and get something more exact?
(1 vote)
• Yes, you can use the left and right-hand values (not limits) and use them to calculate a somewhat more accurate approximation. This is called the trapezoid method, and Sal covers it later in this sequence.