[Ali73]

Use the following table of values to estimate the area under the graph of f(x) over [0, 1] by computing the average of R5 and L5.

x 0 0.2 0.4 0.6 0.8 1
f(x) 52 48 46 44 42 37

[Ali73]

Evaluate the limit.
lim(N--> Infinity) RN, f(x)=9x, [0,1]

[Ali73]

∫[0,4,|2x-4|,x]

[Ali73]

f(x) = x3 + 8

[Ali73]

Evaluate the indefinite integral
∫(8/x+8e^(x))*dx

[Ali73]

Find a formula for RN for the given function and interval. Then compute the area under the graph as a limit.
f(x) = x3 + 3x2, [0, 3]

[ebaines]

Do you mean this:



You need to calculate the definite integral:



Do you know how to do that? I'm not sure what you mean by "RN" - please clarify.

[Curlyben]

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[liongal]

LMAO... That's so funny, but I bet it was a homework q as well

[jcaron2]

I'm not familiar with the L5/R5 nomenclature either, but I think you're supposed to estimate the area under the curve by pretending that it's comprised of rectangles at the specified heights. I think the "L" and "R" mean "left" and "right", where the designation refers to whether the y-value forms the left or right corner of the rectangle. I'm not sure what the "5" means, other than the curve will be divided into five intervals (as specified by the six six points on the curve).

So in the "L" representation, for example, the first rectangle (from x=0 to x=0.2) would have a height of 52 (and therefore an area of 10.4) because its height is determined by the y-value on it's left, and the fifth rectangle (from x=0.8 to x=1) would have a height of 42 (area = 8.4). In the "R" representation, on the other hand, the first rectangle would have a height of 48 (area = 9.6), and the last would have a height of 37 (area = 7.4).

Neither the "L" or the "R" are very accurate at estimating the area. The "L" representation will tend to underestimate when the absolute value of the function is increasing (i.e. when the function is growing further from the x-axis in either the up or down direction as you move to the right), and overestimate when the absolute value is decreasing. The "R" version is just the opposite.

I think the problem is asking you to compute both the "L" and "R" estimates and average the two together (which should, indeed, be a better estimate than either one by itself). It's simply a matter of computing all five rectangular areas for each of the two representations (two of each of which are already done above), adding them up, and dividing by two.