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Calculus/Double Integral question

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Expert: - 12/8/2008


Question
Here is my problem:  
Calculus: evaluate ∫(INT 9 to 8) ∫(INT 7y to 5y) (7/x) dx dy

I would like to see it worked all the way out because I'm having a crazy time for some reason with the algebra at the end as well as the integration. Thanks alot.

Answer
When we are integrating 7/x, the 7 can be put out front so we are only integrating 1/x.  From calculus, the integral of 1/x is ln(x).
Evaluating this from 7y down to 5y gives ln(7y) - ln(5y).

When two ln() functions are subtracted, it's the same as division.
I mean ln(a) - ln(b) = ln(a/b).

Applying this, we have

⌠9
⌡8 ln(7y/5y) dy =

⌠9
⌡8 ln(7/5) dy =

ln(7/5)y from 8 to 9, or ln(7/5)(9-8) = ln(7/5).

Now at the start, we put the 7 out front.  Bringing that back in gives 7ln(7/5).  Remember that aln(x) = ln(x^a), so 7ln(7/5) =
ln(7^7/5^7) = ln(823,543/78,125), which is approximately
ln(10.5413504).  

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