For those functions whose integrals we can't find exactly, we can still use the left-hand sum, right-hand sum, midpoint sum and trapezoid sum to estimate their integrals.
The left-hand, right-hand, and midpoint sums are examples of Riemann Sums.
A Riemann Sum is any sum you get where you split up [a,b] into sub-intervals. The intervals don't necessarily all of the same size. Draw a rectangle on the sub-interval using a value of the function on that sub-interval for the rectangle "height."
We put "height" in quotes because now the functions are allowed to take on negative values. While we can't have negative heights, we can think of function value as a weighted height. If the weighted height is negative and the sub-interval goes from left to right, then the weighted area will also be negative.
Whether the LHS, RHS, MID, and TRAP are over- or under-estimates doesn't depend at all on whether f is positive or negative.
Sample Problem
Let f be negative and increasing. The right-hand sum rectangles don't cover enough area, but when f is negative this means the right-hand sum will be less negative than the actual integral. Therefore the right-hand sum will be an over-estimate.
In order to determine if your sum gives an over-or under-estimate, you can rely on the things we learned earlier:
- If f is increasing then the left-hand sum is an over-estimate and the right-hand sum is an under-estimate.
- If f is decreasing then the left-hand sum is an under-estimate and the right-hand sum is an over-estimate.
- If f is concave up then the trapezoid sum is an over-estimate and the midpoint sum is an under-estimate.
- If f is concave down then the trapezoid sum is an under-estimate and the midpoint sum is an over-estimate.
It doesn't matter for any of these if f is positive or negative. Also, remember that you don't need to remember all of this. You can get by with this list and knowing how to draw graphs.
Practice:
Let f (x) = x3. Use a left-hand sum with 3 sub-intervals to estimate . Is your answer an over- or under-estimate? | |
We use exactly the same process as we did for the non-negative functions, except now some of our "heights" will be negative. First, split up the interval [-3,0] into 3 sub-intervals of size 1. On each sub-interval, find the value of f at the left-endpoint of the sub-interval and use it to draw a rectangle. Multiply each value of f by the length of its sub-interval, and add the results: f (-3)(1) + f (-2)(1) + f (-1)(1) = (-27)(1) + (-8)(1) + (-1)(1) = -36. Or, do the shortcut: add the values of f, then multiply by the length of a sub-interval: [f (-3) + f (-2) + f (-1)](1) = [-27-8-1](1) = -36. We estimate 
The left-hand sum rectangles cover more area than we would like. However, since this area is all below the x-axis, the left-hand sum gives us a more negative value than the actual integral. This means the left-hand sum is an underestimate. | |
Use a midpoint sum with 2 sub-intervals to estimate . Is your answer an over-estimate or an under-estimate? | |
We split the interval [-4,4] into two sub-intervals of length 4 and find the midpoint of each. We draw rectangles using the values f (-2) = -4 and f ( 2 ) = -4. We add the values of the rectangles and get 4(4) + 4(4) = 32. This is an over-estimate, because f is concave down.
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Use LHS(3) to approximate . Is your answer an over- or under-estimate? | |
We split the interval [-2,4] into three sub-intervals of length 2. On [-2,0] we draw a rectangle with "height" f (-2) = -9 and weighted area -9(2) = -18. On [0,2] we draw a rectangle with "height" f ( 0 ) = -3 and weighted area (-3)(2) = -6. On [2,4] we draw a rectangle with height f ( 2 ) = 3 and area 3(2) = 6. Combining the weighted areas, we get LHS(3) = 8-18-6 = -18. This is an underestimate, because f is increasing and we're using a left-hand sum. | |
Use TRAP(1) to estimate . | |
This one's kind of weird. When we connect the points on the function at the endpoints of the interval [-2,1], instead of getting something that looks like a trapezoid, we get a piece below the x-axis and a piece above the x-axis. Thankfully, we don't need to think about this too hard; we can use the same formula as before. To get the weighted area of this "trapezoid," we average the values of f at the endpoints and multiply by the width of the interval: 
If you look more closely at the picture, you'll see that the line we get from drawing the "trapezoid" forms one big triangle below the x-axis, and one little triangle above the x-axis. If you take the area of the little triangle and subtract the area of the big triangle, you should get -10.5 again. | |
For the integral, (a) find the indicated approximation, and(b) determine if your approximation is an under-estimate, an over-estimate, or exact.
- RHS(3) for

Answer
- We divide the interval [-5,1] into 3 sub-intervals of length 2. On each sub-interval we use the value of f at the right endpoint as the "height" of the rectangle.

- Since the function we're integrating is increasing, a right-hand sum is an over-estimate.
For the integral, (a) find the indicated approximation, and(b) determine if your approximation is an under-estimate, an over-estimate, or exact.
- LHS(5) for

Answer
- We divide the interval [-10,0] into 5 sub-intervals of length 2. On each sub-interval we use the value of f at the left endpoint as the "height".
- Since the function we're integrating is decreasing, a left-hand sum is an overestimate.
For the integral, (a) find the indicated approximation, and(b) determine if your approximation is an under-estimate, an over-estimate, or exact.
- MID(3) for

Answer
- We split the interval [-4,2] into sub-intervals [-4,-2], [-2,0], and [0,2], which have midpoints x = -3,-1,1 respectively. All sub-intervals have length 2.
- Since the function we're integrating is concave up, the midpoint sum is an underestimate.
For the integral, (a) find the indicated approximation, and(b) determine if your approximation is an under-estimate, an over-estimate, or exact.
- TRAP(5) for

Answer
- We divide the interval [-10,0] into 5 sub-intervals of length 2. We find the value of f at each endpoint and draw the trapezoids.
- To find the actual trapezoid sum, we can use our handy formula.

- Since the function we're integrating is concave down, the trapezoid sum is an underestimate.
For the integral, (a) find the indicated approximation, and(b) determine if your approximation is an under-estimate, an over-estimate, or exact.
- TRAP(4) for

Answer
- This time we get sub-intervals of length 5. We can see our answer to (c) from the picture.
- Since a straight line has no concavity, the trapezoid rule gives the exact value of the integral. We can verify this by finding the areas of the triangles above and below the x-axis and calculating the value of the integral exactly. The line
crosses the x-axis when x = 6. The big triangle has base 16 and "height" f (-10) = -8, so its weighted area is -64.
The little triangle has base 4 and height f (10) = 2, so its weighted area is 4. Combining these, we get

which confirms that the trapezoid rule gave us the exact value of the integral.