I got this question and im having trouble with, how do i go about doing it.\

question 7

http://users.bigpond.net.au/sterps/p...s/integral.JPG

and also

question 10

http://users.bigpond.net.au/sterps/p.../integral2.JPG

thanks

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- Mar 24th 2006, 04:57 PMsterpsImproper integral question
I got this question and im having trouble with, how do i go about doing it.\

question 7

http://users.bigpond.net.au/sterps/p...s/integral.JPG

and also

question 10

http://users.bigpond.net.au/sterps/p.../integral2.JPG

thanks - Mar 25th 2006, 12:39 AMRich B.
Hi:

As the present time is 2:14 AM, I will get you started with the first problem, and return in a few hours to finish up (hopefully).

First, we interpret the improper integral, i.e., int[f(x) dx] ; [a,inf], as the limit of the*definite integral*on interval [a,n] as n --> inf. In the present case the lower limit of integration is 3, making the interval [3,n].

Now, if we complete the square on denominator x^2+2x+10, the result is (x^2+2x+1) + 9, or (x+1)^2 + 9. So we have:

Lim_n->inf [12 int (dx / ((x+1)^2 + 9))]. Yes, I know it's time to learn latex.

Letting x+1 = 3tan(ß) gives dx = 3sec^2(ß) * dß. Hence the integral becomes:

12*3 int [sec^2(ß) * dß / (9tan^2(ß) + 9)] = 4 int [sec^2(ß) * dß / (tan^2(ß) + 1)].

Applying the identity tan^2(ß) + 1 = sec^2(ß) leaves 4 int [sec^2(ß) * dß / sec^2(ß) ] = 4 int (dß) = 4ß.

Now, recall the substitution x+1 = 3tan(ß). Because the interval on x is [3,n], we can derive an equivalent interval on ß by solving for ß in,

1) 3+1 = 3tan(ß) and 2) n+1 = 3tan(ß). Thus the new interval is [arctan(4/3), arctan((n+1)/3)]. And the integral, i.e., 4ß ; [x=(3,n)] is evaluated as:

4ß ; [arctan(4/3), arctan((n+1)/3)] = 4 * [arctan((n+1)/3) - arctan(4/3)].

I leave the finishing details to you. I'm off to catch up on some sleep.

Regards,

Rich B.