Finding the area of a region between curve and horizontal axis

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SUMMARY

The discussion focuses on calculating the area under the curve defined by the equation y = 6x³ - 2 for the interval 5 ≤ x ≤ 10. Participants debated the appropriate methods, including sigma notation and Riemann sums, to approach the problem. It was clarified that the correct lower boundary for T0 is 5, and the sum of i³ from 1 to n is given by [n(n+1)/2]². Ultimately, the use of a calculator to find the antiderivative was recommended, leading to consistent results.

PREREQUISITES
  • Understanding of Riemann sums
  • Familiarity with sigma notation
  • Knowledge of definite integrals
  • Basic calculus concepts, including antiderivatives
NEXT STEPS
  • Learn about the properties of Riemann sums in calculus
  • Study the process of finding antiderivatives using integration techniques
  • Explore the application of sigma notation in calculus problems
  • Practice calculating areas under curves using both analytical and numerical methods
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Students studying calculus, particularly those focusing on integration techniques and area calculations under curves, as well as educators looking for examples of problem-solving in mathematical discussions.

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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