Integral of 2/(Y+1) - Solution without Calculator

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Discussion Overview

The discussion revolves around finding the integral of the function 2/(Y+1) over the interval [2,0]. Participants explore various methods for calculating the integral analytically without the use of a calculator, focusing on antiderivatives and substitution techniques.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to compute the integral of 2/(Y+1) without a calculator, noting their familiarity with the antiderivative of (x)^-1 being ln(x).
  • Another participant suggests integrating by recognizing that the antiderivative of 1/(Y+1) is ln(Y+1) and proposes a substitution method to evaluate the integral.
  • A third participant reinforces the idea that the antiderivative of 1/(Ax+B) is ln(A + Bx)/A, suggesting a change of variable to simplify the integration process.
  • One participant provides a specific calculation, arriving at the result of 3ln3, while expressing hope that their method is correct.
  • Another participant later claims that the correct answer is 2ln3, indicating a potential disagreement or correction regarding the previous calculations.

Areas of Agreement / Disagreement

There is no consensus on the final answer, as participants propose different results (3ln3 vs. 2ln3) and methods for solving the integral. The discussion remains unresolved regarding the correct approach and outcome.

Contextual Notes

Participants mention the need to change the limits of integration when performing variable substitutions, but the specifics of these changes are not fully explored. There is also a lack of clarity on the assumptions made in the calculations presented.

Whitebread
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Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?
 
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Whitebread said:
Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?

well y' = 2/(Y+1) remember that

Integrate:
2/(Y+1) * y'/y'
2/y' * ln(y+1)
substitute y'
(y+1)ln(y+1)|2,0
[3ln3 - 0]
3ln3 is the answer
hope i did it right...
 
Whitebread said:
Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?


well, clearly if you know that the antiderivative of 1/x is ln (x) then the antiderivative of 1/(x+1) is ln (1 +x) (check! Take the derivative of the result!). and the antiderivative of 1/(Ax+B) with A and B being constants is ln(A +B x) /A (check!).

The way to prove it is to simply do the obvious change of variable, u = 1+x, dx=du and then integrate over u, which you know how to do. (If you do this change of variable to do yoru problem, don't forget to change the limits of integration.)
 
Last edited:
Remember that dx=d(x+c), c=const.

- Kamataat
 
Hi guys,
the correct answer is 2ln3, thanks for the help guys!
 

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