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

  • Thread starter Triggy
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In summary, the given expression evaluates to 0 because it represents the area enclosed between the curve and the x-axis between the values of 1 and squareroot3. This can be simplified to "g(squareroot3)-g(squareroot3)" which is equal to 0. However, there seems to be some confusion regarding the value of \arccos{\sqrt{3}}, with Werg22 pointing out that the problem makes no sense as a real valued integral since \sqrt{3} is larger than 1. It is suggested that Triggy may have made a typo and meant either \int^{1/\sqrt{3}}_{1/\sqrt{3}} or arctan instead of
  • #1
Triggy
8
0
Why does
[tex] \int^{1\sqrt3}_{1\sqrt3} ^5{\sqrt{x^2 + arccos{x}}}dx [/tex]

= 0
 
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  • #2
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)"
 
  • #3
And just what is [tex]\arccos {\sqrt{3}} [/tex]?
 
  • #4
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!
 
  • #5
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]
 
  • #6
either that or arctan instead of arccos but i highly doubt it seeing it is 1squareroot3 not just squareroot3
 

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

What is the integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3?

The integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3 is equal to 0. This is because the function is symmetrical about the y-axis, so the area above the x-axis is cancelled out by the area below the x-axis.

How do you solve for the integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3?

To solve for this integral, you can use the substitution method. Let u = x^2+arccos(x), then du = (2x-sin(arccos(x)))dx. This simplifies the integral to ∫Sqrt(u)du, which can then be solved using the power rule for integrals.

What is the significance of the limits in the integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3?

The limits, 1√3 and 1√3, represent the range of values for which the integral is being calculated. In this case, the limits correspond to the points where the function intersects the x-axis, creating a symmetrical interval.

Can the integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3 be approximated using numerical methods?

Yes, the integral can be approximated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods divide the interval into smaller subintervals and use the function values at the endpoints of each subinterval to estimate the area under the curve.

What is the physical interpretation of the integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3?

The integral of Sqrt(x^2+arccos(x)) from 1√3 to 1√3 represents the total distance travelled by an object moving along the x-axis with a velocity described by the given function. It can also be thought of as the total displacement of the object over the given interval.

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