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10 0.8 0.6 04 0.2 -1.5 -1.0 -0.5 0.5 1.0 1.5 where the picture on the left is a graph of e- and on the right is e-(a²+v°). If you view the second figure as a solid, the cross-sections are in fact circles, and you can treat it as a solid of revolution! (4) In other words, rotate the area under e-a around the y-axis, and show that the volume of the solid of revolution is r. Use this to conclude the value of I.

10 0.8 0.6 04 0.2 -1.5 -1.0 -0.5 0.5 1.0 1.5 where the picture on the left is a graph of e- and on the right is e-(a²+v°). If you view the second figure as a solid, the cross-sections are in fact circles, and you can treat it as a solid of revolution! (4) In other words, rotate the area under e-a around the y-axis, and show that the volume of the solid of revolution is r. Use this to conclude the value of I.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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10 0.8 0.6 0.4 0.2 -1.5 -1.0 -0.5 0.5 1.0 15 where the picture on the left is a graph of e-a² and on the right is e-(*²+v°). If you view the second figure as a solid, the cross-sections are in fact circles, and you can treat it as a solid of revolution! (4) In other words, rotate the area under e-a around the y-axis, and show that the volume of the solid of revolution is T. Use this to conclude the value of I.
Transcribed Image Text:10 0.8 0.6 0.4 0.2 -1.5 -1.0 -0.5 0.5 1.0 15 where the picture on the left is a graph of e-a² and on the right is e-(*²+v°). If you view the second figure as a solid, the cross-sections are in fact circles, and you can treat it as a solid of revolution! (4) In other words, rotate the area under e-a around the y-axis, and show that the volume of the solid of revolution is T. Use this to conclude the value of I.
In this problem we will cover some basic probability theory as an application of integration. We will consider a random variable X, for example the height of a randomly chosen person in our class. Given real numbers a, b, there is a function f called the probability density function such that the probability that X is in the interval (a, b) is equal to P(a < X < b) = | s(12)dæ. The hydrogen atom is composed of one proton in the nucleus and one electron, which moves about the nucleus. In the quantum theory of atomic structure, it is assumed that the electron does not move in a well-defined orbit. Instead, it occupies a state known as an orbital, which may be thought of as a “cloud" of negative charge surrounding the nucleus. At the state of lowest energy, called the ground state, or 1s-orbital, the shape of this cloud is assumed to be a sphere centered at the nucleus. This sphere is described in terms of the probability density Sa?e-2#/a0, æ 2 0 p(x) x <0 where ao is the Bohr radius. The integral P(0 < X <r) = | p(x)dx gives the probability that the electron will be found within the sphere of radius r meters centered at the nucleus. (1) Evaluate the limit lim p(r) (you must use l'Hopital's rule). (2) Determine the value of a where p(x) attains its maximum value. (3) Find the probability that the electron will be within the sphere of radius 4ao centered at the nucleus. Now most random phenomena are modeled by what we call the normal, or Gaussian distribution, of which the simplest probability density function is related to the function e-. As it turns out, integrating this function properly requires Calc 3 techniques, but we can use a trick to get it now. We are looking for the value of the integral ´dx. The trick is to square it and treat the second integral as being an different vari- able, "dy -2²-v² dx dy, and as you can see this is a two-variable integral, which of course will have to wait till Calc 3. But it turns out that the picture that you get is the following:
Transcribed Image Text:In this problem we will cover some basic probability theory as an application of integration. We will consider a random variable X, for example the height of a randomly chosen person in our class. Given real numbers a, b, there is a function f called the probability density function such that the probability that X is in the interval (a, b) is equal to P(a < X < b) = | s(12)dæ. The hydrogen atom is composed of one proton in the nucleus and one electron, which moves about the nucleus. In the quantum theory of atomic structure, it is assumed that the electron does not move in a well-defined orbit. Instead, it occupies a state known as an orbital, which may be thought of as a “cloud" of negative charge surrounding the nucleus. At the state of lowest energy, called the ground state, or 1s-orbital, the shape of this cloud is assumed to be a sphere centered at the nucleus. This sphere is described in terms of the probability density Sa?e-2#/a0, æ 2 0 p(x) x <0 where ao is the Bohr radius. The integral P(0 < X <r) = | p(x)dx gives the probability that the electron will be found within the sphere of radius r meters centered at the nucleus. (1) Evaluate the limit lim p(r) (you must use l'Hopital's rule). (2) Determine the value of a where p(x) attains its maximum value. (3) Find the probability that the electron will be within the sphere of radius 4ao centered at the nucleus. Now most random phenomena are modeled by what we call the normal, or Gaussian distribution, of which the simplest probability density function is related to the function e-. As it turns out, integrating this function properly requires Calc 3 techniques, but we can use a trick to get it now. We are looking for the value of the integral ´dx. The trick is to square it and treat the second integral as being an different vari- able, "dy -2²-v² dx dy, and as you can see this is a two-variable integral, which of course will have to wait till Calc 3. But it turns out that the picture that you get is the following:
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