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.
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
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.
Oh! I understand now. The "total area" part kind of threw me off.
Thank you so much!
To find the total area under the curve for x ≥ 1, we need to calculate the definite integral of the curve from 1 to infinity. However, since the curve is asymptotic to the x-axis at x = 0, we need to be careful when evaluating the integral.
The indefinite integral of y = 1/(x^3) is -1/(2x^2), so the definite integral from 1 to t is given by:
∫[1 to t] (1/(x^3)) dx = [-1/(2x^2)] evaluated from 1 to t
= [-1/(2t^2)] - [-1/(2(1)^2)]
= [-1/(2t^2)] - (-1/2)
= (-1/(2t^2)) + (1/2)
= (1/2) - [1/(2t^2)]
You have already correctly found this result.
Now, to find the total area under the curve for x ≥ 1, we need to take the limit of the definite integral as t approaches infinity:
lim[t->∞] ((1/2) - [1/(2t^2)])
As t approaches infinity, the term [1/(2t^2)] goes to zero since the value of t^2 in the denominator increases much faster than the numerator. Therefore, the limit simplifies to:
lim[t->∞] ((1/2) - [1/(2t^2)]) = 1/2
Hence, the total area under the curve for x ≥ 1 is 1/2.
In summary, the area of the curve for x = 1 to x = t is (1/2) - [1/(2t^2)], and the total area under the curve for x ≥ 1 is 1/2.