Integral of a function with a square root in denominator

In summary, the integral of (x dx)/sqrt(1-x^2) on the interval [2,5] is equal to 2.863. However, upon checking with a calculator, the correct answer is 0.716. The error was due to dividing u^(1/2) by 2 instead of 1/2. The correct solution involves using the equation integral u^n(du) = u^(n+1)/(n+1) and changing the limits of integration after making the change of variables.
  • #1
judahs_lion
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Homework Statement


integrate (xdx)/sqrt(1-x^2) on the integreal [2,5]


Homework Equations


integral u^n(du) = u^n+1/(n+1)


The Attempt at a Solution


I'm no good at using latex so I scanned in my work.

I get .716 but when I check with my calculator it comes out as 2.863
 

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  • #2


I see where I messed up. I was dividing u^1/2 by 2 instead of 1/2. Thanks anyways.
 
  • #3


Two things:

[tex]\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.[/tex]

Also you forgot to change the limits of integration after making the change of variables.
 
  • #4


fzero said:
Two things:

[tex]\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.[/tex]

Also you forgot to change the limits of integration after making the change of variables.

Actually I think I did do that. Not sure if u can see on the side where i did u(2) =5 and u(5) = 26.
 

What is the integral of a function with a square root in the denominator?

The integral of a function with a square root in the denominator is a mathematical concept that involves finding the area under the curve of the function. It is denoted by ∫f(x)dx and is a fundamental concept in calculus.

Why is it important to find the integral of a function with a square root in the denominator?

Finding the integral of a function with a square root in the denominator is important in many applications, such as calculating areas and volumes, determining the average value of a function, and solving differential equations. It also helps in evaluating complex mathematical expressions and understanding the behavior of a function.

What is the process for finding the integral of a function with a square root in the denominator?

The process for finding the integral of a function with a square root in the denominator involves using substitution or integration by parts. In some cases, it may also require the use of trigonometric identities or other integration techniques. It is important to have a good understanding of basic calculus principles and techniques in order to successfully find the integral of such functions.

What are some common mistakes that people make when finding the integral of a function with a square root in the denominator?

One common mistake is forgetting to use the chain rule when integrating by substitution. Another mistake is not recognizing when to use trigonometric identities or other integration techniques. It is also important to pay attention to the limits of integration and to check for errors in calculations, as even small mistakes can greatly affect the final result.

Are there any tips or tricks for finding the integral of a function with a square root in the denominator?

Some tips for finding the integral of a function with a square root in the denominator include carefully choosing the substitution variable, checking for common trigonometric identities and integration formulas, and practicing with a variety of examples. It is also helpful to break down the function into simpler parts and use the linearity property of integration. Lastly, double-checking the answer and understanding the concept of integration can also be beneficial in solving these types of integrals.

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