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Mathematics 18 Online
OpenStudy (anonymous):

Using Reimann Sums evaluate ∫1_3{(x^2+2x-1)}dx

OpenStudy (cruffo):

\[\int_1^3 (x^2 + 2x -1)\;dx\] using the left endpoint, right endpoint, or averaged? How many subdivisions should be made?

OpenStudy (anonymous):

either or left or right

OpenStudy (cruffo):

How do you want to divide it up? One big rectangle, two, three, ...?

OpenStudy (anonymous):

using summation...

OpenStudy (cruffo):

hummm.....

OpenStudy (anonymous):

\[\sum_{k=1}^{n}\]

OpenStudy (cruffo):

(nice LaTeX) usually the problems will give you a value for n.

OpenStudy (anonymous):

truthfully i am not sure how to solve this problem so do it the best way you know...and i'll see if i could make sense of it

OpenStudy (cruffo):

For example if we do left summation, and use n = 2, then each rectangle would have a length of (3-1)/2 = 1, so the sum would be f(1) + f(2) If n = 4, then each rectangle would have a length of (3-1)/4 = 0.5, so the left sum would be 0.5f(1) + 0.5f(1.5)+0.5f(2) + 0.5f(2.5)

OpenStudy (anonymous):

wow okay...is that the final answer? thanks so much i have a super migrane and the computer light is making it worse...

OpenStudy (cruffo):

If we use n = 4 and left sum: \[f(1) = (1)^2 + 2(1)-1 = 2\] \[f(1.5) = (1.5)^2 + 2(1.5)-1 = 4.25\] \[f(2) = (2)^2 + 2(2)-1 = 7\] \[f(2.5) = (2.5)^2 + 2(2.5)-1 = 10.25\] So the answer is 0.5(2 + 4.25 + 7 + 10.25) = 11.75 See attached pic for illustration. We are summing up the areas of the shaded rectangles. The larger the n, the more rectangles are used in the sum, so the approximation would be better.

OpenStudy (anonymous):

i got 20/3 using the right-endpoints

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