What is limit of integral f(x)dx, from -n to n?  integral of 1/x^2+x+1

Asked on by thales

1 Answer | Add Yours

Top Answer

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to calculate the limit of the definite integral such that:

`lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1)`

`` You need to complete the square at denominator to make the following basic integral `int dx/(x^2 + a^2).`

`int dx/(x^2+x+1) = int dx/((x^2+2*(1/2)*x+ + 1/4) - 1/4 + 1)`

`int dx/(x^2+x+1) = int dx/((x + 1/2)^2 + (sqrt3/2)^2)`

`int dx/(x^2+x+1) = (2/sqrt3)*arctan((x + 1/2)/(sqrt3/2)) + c`

Evaluating the definite integral yields:

`int_(-n)^n dx/(x^2+x+1) = (2/sqrt3)*(arctan((n + 1/2)/(sqrt3/2)) - arctan((-n + 1/2)/(sqrt3/2)))`

Evaluate the limit:

`lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1) = (2/sqrt3)*(lim_(n-gtoo)arctan((n + 1/2)/(sqrt3/2)) - lim_(n-gtoo) arctan((-n + 1/2)/(sqrt3/2))) =gt`

`=gt lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1) = (2/sqrt3)*((2/sqrt3)*(pi/2) -(2/sqrt3)*(-pi/2))`

`=gt lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1) = (4/3)*(2pi/2)`

`lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1) = (4/3)*(pi)`

Evaluating the limit of the definite integral yields:`lim_(n-gtoo) int_(-n)^n dx/(x^2+x+1) = (4/3)*(pi).`

We’ve answered 319,865 questions. We can answer yours, too.

Ask a question