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S(1/(x^2 + 1) - C/(3x+1)) dx from 0 to inifinity  Find the value of the constant C...

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tgl223 | Student, College Freshman | eNoter

Posted February 27, 2011 at 6:08 AM via web

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S(1/(x^2 + 1) - C/(3x+1)) dx from 0 to inifinity

 

Find the value of the constant C for which the integral converges. Evaluate the integral for this value of C.

Essential Calculus Early Transcendentals by James Stewart

Chapter 6.6 Q#62

Tagged with improper integral, math

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 21, 2013 at 8:11 AM (Answer #1)

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Since the upper endpoint is infinite, you need to evaluate the improper integral such that:

`int_0^(oo)(1/(x^2 + 1) - C/(3x + 1)) dx = lim_(n->oo) int_0^n(1/(x^2 + 1) - C/(3x + 1)) dx`

`lim_(n->oo) int_0^n(1/(x^2 + 1) - C/(3x + 1)) dx = lim_(n->oo)` `int_0^n 1/(x^2 + 1) dx - lim_(n->oo) int_0^n C/(3x + 1) dx`

Evaluating the definite integral `int_0^n 1/(x^2 + 1) dx` yields:

`int_0^n 1/(x^2 + 1) dx = tan^(-1) x |_0^n`

Using he fundamental theorem of calculus yields:

`int_0^n 1/(x^2 + 1) dx = tan^(-1) n - tan^(-1) 0`

`int_0^n 1/(x^2 + 1) dx = tan^(-1) n `

Evaluating the definite integral `int_0^n C/(3x + 1) dx` yields:

`int_0^n C/(3x + 1) dx = C*ln|3x + 1||_0^n`

`int_0^n C/(3x + 1) dx = C*(ln|3n + 1| - ln 1)`

Since `ln 1 = 0` yields:

`int_0^n C/(3x + 1) dx = C*(ln|3n + 1|)`

Replasing the results `tan^(-1) n` and `C*(ln|3n + 1|)` under limits, yields:

`lim_(n->oo) int_0^n 1/(x^2 + 1) dx - lim_(n->oo) int_0^n C/(3x + 1) dx = lim_(n->oo) tan^(-1) n - lim_(n->oo) C*(ln|3n + 1|)`

`lim_(n->oo) int_0^n 1/(x^2 + 1) dx - lim_(n->oo) int_0^n C/(3x + 1) dx = pi/2 - lim_(n->oo) ln(3n + 1)^C`

Since the integral converges, hence, the limit `lim_(n->oo) ln(3n + 1)^C` needs to be finite, such that:

`lim_(n->oo) C*(ln|3n + 1|) = lim_(n->oo) C*(ln n*(3 + 1/n)) `

`lim_(n->oo) C*(ln n) + lim_(n->oo) C (3 + 1/n) = lim_(n->oo) C*(ln n) + 3C`

If `C = 0` , hence, `lim_(n->oo) C*(ln n) = lim_(n->oo) ln n^C=> lim_(n->oo) ln n^0 = lim_(n->oo) ln 1 = 0`

Hence, evaluating C, solving the given improper integral, under the given conditions, yields `C = 0.`

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