Expert: Sherry Wallin - 12/8/2008

Question
I know this shouldn't be that much harder than the double integral problems with fixed number intervals but I seems to be getting lost in all the clutter as I attempt to work through these problems. I think if I saw one worked out I could figure what I'm doing wrong.

Calculus: evaluate ∫(INT 9 to 8) ∫(INT 7y to 5y) (7/x)dx dy?

Thanks so much,
Shannon

Answer
Hi Shannon~
    This is a very "messy" problem, but let's try getting through it without making any careless errors...
∫(INT 9 to 8) ∫(INT 7y to 5y) (7/x)dx dy
let's rewrite 7/x ad 7x-1
∫(INT 9 to 8) ∫(INT 7y to 5y) (7x^-1)dx dy
∫(INT 9 to 8)  ((7/-2)x^-2)(evaluated between 7y to 5y) dy
(-7/2)∫(INT 9 to 8)(1/x^2)(evaluated between 7y to 5y) dy
(-7/2)∫(INT 9 to 8){[1/(7y)^2]-[1/(5y)^2]}dy
(-7/2)∫(INT 9 to 8)[1/(49y^2)]-[1/(25y^2)]dy
(-7/2)((1/49)-(1/25))∫(INT 9 to 8)(1/y^2)dy
(-7/2)((1/49)-(1/25))(-1/3)(1/y^3)(evaluated from 9 to 8)

(-7/2)((1/49)-(1/25))(-1/3)[(1/9^3)-(1/8^3)]

I think you can do the rest, it's just arithmetic. If you have any
questions about any of the steps please ask for clarification.

Math Prof