Integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3

  • Context: Graduate 
  • Thread starter Thread starter Triggy
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Discussion Overview

The discussion revolves around the integral of the function \(\sqrt{x^2 + \arccos{x}}\) evaluated from \(1/\sqrt{3}\) to \(1/\sqrt{3}\). Participants explore the implications of this integral, particularly focusing on its value and the conditions under which it is defined.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that the integral evaluates to zero because it represents the area under the curve between the same limits, leading to the expression \(g(\sqrt{3}) - g(\sqrt{3})\).
  • Another participant emphasizes that the integral \(\int_a^a f(x)dx\) equals zero for any integrable function \(f\), provided that \(F\) is an anti-derivative of \(f\).
  • Werg22 raises a concern that since \(\sqrt{3}\) is greater than 1, the integral may not make sense as a real-valued integral.
  • There is speculation about a potential typo in the original post, suggesting that the limits might have been intended as \(1/\sqrt{3}\) instead of \(1\sqrt{3}\).
  • Another participant questions whether \(\arccos\) was the intended function, proposing that \(\arctan\) might have been meant instead, though they express doubt about this interpretation.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the integral as stated, with some suggesting it is nonsensical due to the limits provided. There is no consensus on whether the original expression was correct or if a typo occurred.

Contextual Notes

The discussion highlights potential ambiguities in the notation used for the limits of integration and the function involved, which may affect the interpretation of the integral.

Triggy
Messages
8
Reaction score
0
Why does
[tex]\int^{1\sqrt3}_{1\sqrt3} ^5{\sqrt{x^2 + arccos{x}}}dx[/tex]

= 0
 
Physics news on Phys.org
because by this you mean find the area enclosed between the curve and the x-axis between squareroot3 and squareroot3. its the integral of it at squareroot3 minus the integral of it at squareroot3. so if we let the integral function be g(x) your expression says "g(squareroot3)-g(squareroot3)"
 
And just what is [tex]\arccos {\sqrt{3}}[/tex]?
 
triggy, itsjustme's point is that
[tex]\int_a^a f(x)dx= F(a)- F(a)= 0[/tex]
for any integrable function f (F is, of course, an anti-derivative of f).

Werg22's point is that, since [itex]\sqrt{3}[/itex] is larger than 1, the problem makes no sense as a real valued integral!
 
HallsofIvy said:
Werg22's point is that, since [itex]\sqrt{3}[/itex] is larger than 1, the problem makes no sense as a real valued integral!

My guess is that Triggy made a typo, i.e., Triggy meant

[tex]\int^{1/\sqrt{3}}_{1/\sqrt{3}}[/tex]
 
either that or arctan instead of arccos but i highly doubt it seeing it is 1squareroot3 not just squareroot3
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K