# 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.

1. 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 x--inf} - lim {(-1/2)/x^2 as x=1
= 0 - (-1/2) = 1/2

2. Reiny

or

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.

3. Alyssa

Oh! I understand now. The "total area" part kind of threw me off.

Thank you so much!

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