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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.
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?
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, then add the values of the rectangles and get -4(4) + -4(4) = -32. Since f is concave down we covered too much area, but since the area is negative we have an underestimate.
Use LHS(3) to approximate . Is your answer an over- or under-estimate?
Wplit 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) = -18 + -6 + 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.