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### Course: Integral Calculus>Unit 1

Lesson 4: Riemann sums in summation notation

# Riemann sums in summation notation: challenge problem

When a function is negative, Riemann sums seem to treat it as having "negative area".

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• Confused me as I thought sum might still mean the total of a positive and negative value
• Although it sounds tempting, area cannot be negative. When a function dips below the x-axis, the space between the function and the x-axis should be treated the same way as if it were above the x-axis. Because there is no absolute value sign, both II and III are wrong as they subtract area from themselves.
• So, can you actually use the Riemann sum at all to work out the area of the rectangles that are in the third/fourth quadrants? Or is it just impossible because you'd never be able to get a positive number?
• You could always introduce the absolute value function into the sum. To say this more precisely, instead of doing the summation of that function, you can do the summation of the absolute value of that function. In a sense, this would force all the values of the function to be positive which makes it give you the height of every rectangle.
• It confused me when Sal ruled out the second choice. Traditionally, you cannot have a negative area, but I thought that area under the x-axis above the curve was considered negative. At least, it is when evaluating signed area.
• The way that areas below the x-axis are interpreted when integrating depends on what the function is representing. In other words, is the function representing something such that any area below the x-axis should be subtracted or added to the area above the x-axis. So, there is not a single answer -- what you do depends on what the area is representing.
• For the part where you discuss whats inside the f(x), how is that related to whether it is negative or positive? its unclear why the statements are true and false
• I was confused about this as well. You have to remember that the f(stuff) is directing you to plug the value of the "stuff" you just found into whatever function you're dealing with. For example, the first red rectangle you would have f( -1 + 1/2 ) which equals f( -1/2 ). When you plug -1/2 into the equation being graphed you get some very small negative value right below the x-axis. That small negative value is what causes the trouble since no matter what red rectangle you're referencing, the corresponding f(stuff) = y = "rectangle height" value is always going to be negative which will give you a negative approximate area. TL;DR a rectangle below the x-axis may be 4 units tall, but the y-value you're using to determine that height is actually negative and that y-value is what is being used in the sum, not the absolute value of y which is the height.
• confuses me why we dont say what we said for second (it can not be negative) also for first one
• We don't take the x-values into consideration when calculating the area, just the distance between them (1/2) which is always positive so it doesn't matter whether the x-values are positive or negative.
We do, however, use the y-values as the height which can be positive or negative and that's where we get into the issue of negative areas. The first example has all positive y-values while the second has all negative y-values and the third has a mixture of both.
• I don't get why it was F(5 + i/2). I know that it is the height, but how did he get that expression?
• It's actually f(-5+i/2) (emphasis on the negative five). It is set up this way because the rectangles in the graph start at x = -5 and increment by 1/2 for each rectangle. Because it's a right-handed rectangle, the y value that determines the rectangle's height is on the right hand side. So the first rectangle's height is given by the y-value f(-5 + 1/2), the second recangle's by f(-5 +2/2), the third by f(-5 + 3/2), and so on. Notice that the expression says "the value of f at negative 5 (where the rectangles start) plus 1/2 times whichever rectangle we're currently on."

If it's the iteration from i = 1 through i = 8 that is confusing, look into videos on sigma notation. That should help clear it up.
• If this graph were said to describe the velocity of an object against time, could it be said that the absolute value of the area (taking the value of the integral between x=-5 and x=-1 and adding it to the absolute value of the integral between x=-1 and x=7) is equal to the total distance travelled, but the value of area (value of the integral between x=-5 and x=7, so subtraction would occur) is equal to the displacement?
• So why can't area be negative?
• Sounds like an unfair question. We might consider negative areas as having a decrease in the antiderivative function.