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


    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.

  3. Alyssa

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

    Thank you so much!

Respond to this Question

First Name

Your Answer

Similar Questions

  1. calculus II

    We're doing areas by integrals now, with 2 eqns. I have a few questions. 1. Sketch the region in the xy-plane defined by the inequalities x-2(y^2)> 0 and 1-x-abs(y)>0. and find its area. Would it be best to solve for x, then …
  2. Calculus

    Integrals: When we solve for area under a curve, we must consider when the curve is under the axis. We would have to split the integral using the zeros that intersect with the axis. Would this be for all integrals?
  3. Math: Need Answer to study for a quizz. Help ASAP

    Estimate the area under the curve f(x)=x^2-4x+5 on [1,3]. Darw the graph and the midpoint rectangles using 8 partitions. Show how to calculate the estimated area by finding the sum of areas of the rectangles. Find the actual area under …
  4. calculus

    Find area of the region under the curve y = 5 x^3 - 9 and above the x-axis, for 3¡Ü x ¡Ü 6. area = ?
  5. math, calculus 2

    Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from …
  6. brief calc

    Calculate the total area of the region described. Do not count area beneath the x-axis as negative. Bounded by the curve y = square root of x the x-axis, and the lines x = 0 and x = 16 This is under Integrals, i don't know what i'm …
  7. Calculus

    Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the …
  8. Calculus

    Suppose the total area under the curve y=x^2-mx+1 from 0 to 5 is 11.3024. Find m. I know that this means that i need to find the integral of the absolute value of x^2-mx+1 from 0 to 5 and set it equal to 11.3024, but I am not sure …
  9. Calculus

    Calculate the Riemann sum of the area under the curve of f(x)=9-x^2 between x=-2 and x=3 The answer I come up with is 10/3, but when I solve using integrals, the answer I get is 100/3. Am I doing something wrong?
  10. Statistics

    Find the area under the normal curve in each of the following cases. 1. Between z = 0 and z = 1.63 2. Between z = 1.56 and = 2.51 3. Between z = -0.76 and z =1.35 4. Between z= -0.26 and z = -1.76 5. To the left of z = 2.35 6 To the …

More Similar Questions