Solving Integral Confusion: Limits & Periods

In summary: The rest of the conversation discusses using the properties of the function and its period to solve the limit. In summary, the given function has a period of pi due to the squared cosine term. The conversation then discusses using properties of the function and its period to solve the limit.
  • #1
Vali
48
0
Hi!

$$\lim_{n \to \infty }\int_{0}^{n}\frac{dx}{1+n^{2}\cos^{2}x }$$

I found to solution on the internet but I didn't understood it 100%.
First, it says that the function under integral has period $pi$.Why pi ? I know that cos function has period $2kpi$
Consequence: $\int_{0}^{k\pi }f_n(x)dx=k\int_{0}^{\pi }f_n(x)dx$ ( I understood this consequence )
Secondly, to solve the limit I used $x-1<[x]\leq x<[x]+1\leq x+1$ so if $I_n=\int_{0}^{\frac{n}{\pi}\cdot \pi }f_n(x)dx$ I have the following inequalities which I didn't understand completely.
$(\frac{n}{\pi }-1)\int_{0}^{\pi }f_n<\left [ \frac{n}{\pi } \right ]\int_{0}^{\pi }f_n=\int_{0}^{\left [ \frac{n}{\pi } \right ]\pi }f_n\leq I_n<\int_{0}^{(\left [ \frac{n}{\pi } \right ]+1)\pi }f_n=(\left [ \frac{n}{\pi } \right ]+1)\int_{0}^{\pi }f_n<(\frac{n}{\pi }+1)\int_{0}^{\pi }f_n$
Here is my main confusion.These inequalities comes from $x-1<[x]\leq x<[x]+1\leq x+1$ so if $I_n=\int_{0}^{\frac{n}{\pi}\cdot \pi }f_n(x)dx$ but who is $x$ in my integral ?
For me x=In, right ? so it should be $I_{n}-1<[I_{n}]\leq I_{n}<[I_{n}]+1\leq I_{n}+1$
$I_{n}$ remain the same like in the first row of inequalities.It doesn't change.
But how $I_{n}-1$ changed to $(\frac{n}{\pi }-1)\int_{0}^{\pi }f_n$ ?
How $[I_{n}]$ changed to $\left [ \frac{n}{\pi } \right ]\int_{0}^{\pi }f_n$ ?
And so on..
I hope you understand what I don't understand.I mean from initial inqualities results that $I_{n}$ should is $x$.What properties have been used here?
Thanks!
 
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  • #2
The given function has period [tex]\pi[/tex] because the cosine is squared. For half its period, [tex]-\pi/2[/tex] to [tex]\pi/2[/tex], cosine is positive. For the other half, [tex]\pi/2[/tex] to [tex]3\pi/2[/tex], cosine is negative. Squaring makes those the same.
 

Related to Solving Integral Confusion: Limits & Periods

1. What are limits in integral calculus?

Limits in integral calculus refer to the values that a function approaches as the independent variable approaches a certain value. In other words, it is the value that a function "approaches" but may not actually reach at a specific point.

2. How do I solve integrals with limits?

To solve integrals with limits, you can use the fundamental theorem of calculus, which states that the integral of a function f(x) from a to b is equal to the difference of the antiderivatives of f(x) evaluated at a and b. In simpler terms, you can find the antiderivative of the function and then plug in the limits to evaluate the integral.

3. What are periods in integral calculus?

In integral calculus, periods refer to the interval over which a function repeats itself. This can be seen in trigonometric functions, where the period is the length of one complete cycle of the function.

4. How do I find the period of a function in integral calculus?

To find the period of a function in integral calculus, you can use the formula T = 2π/b, where T is the period and b is the coefficient of the independent variable in the function. For example, if the function is y = sin(3x), the period would be T = 2π/3.

5. What are some common mistakes when solving integrals with limits and periods?

Some common mistakes when solving integrals with limits and periods include forgetting to evaluate the antiderivative at the limits, using the wrong limits, and not taking into account the period when evaluating the integral. It is important to double check your work and make sure all steps are followed correctly when solving these types of integrals.

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