Calculus ll  Improper Integrals
posted by Alyssa .
Find the area of the curve y = 1/(x^3) from x = 1 to x = t and evaluate it for t = 10, 100, and 1000. Then find the the total area under this curve for x ≥ 1.
I'm not sure how to do the last part of question ("find the the total area under this curve for x ≥ 1.")
For the area of the curve, I found that the integral from 1 to t is (1/2)[1/(2t^2)].
I used the equation I found and substituted t for 10, 100, and 1000, and got 0.495, 0.49995, and 0.4999995 respectively.

Calculus ll  Improper Integrals 
Reiny
area = LIM [integral]1/x^3 dx from x = 1 to x  infin.
= lim ((1/2)(1/x^2) from x=1 to x inf.
= lim{ (1/2)/x^2 as xinf}  lim {(1/2)/x^2 as x=1
= 0  (1/2) = 1/2 
Calculus ll  Improper Integrals 
Reiny
or
look at your expression of
1/2  1/(2t^2)
as t > infinity, doesn't 1/(2t^2) approach 0 ?
so you are left with 1/2 0
= 1/2 , as your approximations suggested. 
Calculus ll  Improper Integrals 
Alyssa
Oh! I understand now. The "total area" part kind of threw me off.
Thank you so much!
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