Finding the area of a region between curve and horizontal axis

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Homework Help Overview

The discussion revolves around finding the area under the curve defined by the function y = 6x³ - 2 over the interval from x = 5 to x = 10. Participants explore different methods for calculating this area, including sigma notation and Riemann sums.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of sigma notation and Riemann sums, questioning the correct setup for the intervals and the application of limits. There are uncertainties about how to simplify expressions and whether certain terms can be factored out.

Discussion Status

Some participants have provided guidance on the correct interpretation of the problem and the setup of the summation, while others express confusion about the simplification process. There is acknowledgment that multiple methods could be applied, including the use of a calculator for the antiderivative.

Contextual Notes

One participant notes that the instructor indicated a calculator should be used for this problem, suggesting that the expected approach may not align with traditional manual methods. There are also references to additional complexities involving other functions like cosine and natural logarithms.

Razael
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Homework Statement



Find the area under y = 6x3 - 2 for 5 <_ x <_ 10

Homework Equations



The Attempt at a Solution



I'm not sure how it wants us to do it; do we use sigma notation or the general right/left-hand sums?

Using sigma notation, delta T = 10-5/n

T0= 5/n (?)
T1= 10/n
Ti = 5i/n

5/n\sum6(5i/n)^3 -2

750/n^4 -2 (Sigma) i^3

I'm not sure if I can bring the 2 out like that, and I'm also not sure what i^3 equals (is it (n(n+1)/2)^3?)
 
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Without a bit more information it's difficult to know how they expect you to find the answer, because there are numerous ways to do so. You could simply take the definite integral, but it looks like (from your work) you're expected to use the limit definition and Riemann Sums.

Your delta t is correct, but t0 for these problems is usually equal to the lower boundary. t1 and your other subintervals would only be correct if the lower boundary was zero. Think about that.

And no, you're not allowed to pull out the -2 like that. Try to take a look at your Sigma rules to see what you can do to simplify the problem once you find the value of f(ti).

The sum of i3, with i = 1 to n is [n(n+1)/2]2.
 
Looking back, a T0 of 5 does make a lot more sense. T1, then, would be 5 +5/n, T2 would be 5+10/n, so Ti would be 5+5i/n?

5/n\sum6(5+5i/n)^3-2

I've started simplifying this and it doesn't look right.
 
Razael said:
I've started simplifying this and it doesn't look right.

Well, what were you trying and what doesn't look right? :)
 
Sorry for the late response.

It just seemed too "long" for a generic textbook problem; asked the instructor, turns out we were supposed to use a calculator on those ones (other included cos and natural logs inside the summation which I had no idea how to pull out).

Used a calculator and the antiderivative, came to the same answer for both. Thank you for helping even though it turned out to be a waste of time.
 

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