/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q23P Evaluate the following integrals... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the following integrals:

(a)∫-3+1(x3-3x2+2x-1)δ(x+2)dx.

(b).∫0∞[cos(3x)+2]δ(x-π)dx

(c)∫_1+1exp(lxl+3)δ(x-2)dx

Short Answer

Expert verified
  • ∫-3+1(x3-3x2+2x-1)δ(x+2)dx=-25
  • ∫0∞[cos(3x)+2]δ(x-Ï€)dx=1
  • ∫_1+1exp(lxl+3)δ(x-2)dx=0

Step by step solution

01

Significance of the integral

In mathematics, an integral lends numerical values to functions to represent concepts like volume, area, and displacement that result from combining infinitesimally small amounts of data. In mathematics, an integral lends numerical values to functions to represent concepts like volume, area, and displacement that result from combining infinitesimally small amounts of data.

02

(a) Evaluating the integral

Let,

∫-3+1x3-3x2+2x-1δx+2dx

According to impulse property,

x+2=0x=-2

In given function, substitutex=-2.

x3-3x2+2x-1=(-2)3-3(-2)2+2(-2)-1=-8-12-4-1=-25

Hence the value of is ∫-3+1(x3-3x2+2x-1)δ(x+2)dx=-25.

03

:(b) Evaluating the integral

Let,

∫0∞cos3x+2δx-πdx

In given function, substitute x=Ï€, x-Ï€=0x=Ï€.

Then,

cos3Ï€+2=-1+2=1

Hence the value of ∫0∞cos3x+2δx-πdxis, 1.

04

(c) Evaluating the integral

Let,

∫-1+exp(l×l+3)δ(x-2)dx

Then,

∫-1+1expl×l+3δx-2dx=0

0 (x=2is outside the domain of integration).

Because x=2 is out of the integration interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the double delta-function potentialV(x)=-α[δx+a+δx-a]Whereand are positive constants

(a) Sketch this potential.

(b) How many bound states does it possess? Find the allowed energies, forα=ħ/maand forα=ħ2/4ma, and sketch the wave functions.

Show that E must be exceed the minimum value of V(x) ,for every normalizable solution to the time independent Schrodinger equation what is classical analog to this statement?

d2Ψdx2=2mh2[V(x)−E]Ψ;

IfE<Vmin thenΨ and its second derivative always have the same sign. Is it normalized?

A particle of mass m is in the ground state of the infinite square well (Evaluation 2.19). Suddenly the well expands to twice its original size – the right wall moving from a to 2a – leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured.

  1. What is the most probable result? What is the probability of getting that result?
  2. What is the next most probable result, and what is its probability?
  3. What is the expectation value of the energy? (Hint: if you find yourself confronted with an infinite series, try another method)

What is the Fourier transform δ(x) ? Using Plancherel’s theorem shows thatδ(x)=12π∫−∞∞eikxdk.

If two (or more) distinct44solutions to the (time-independent) Schrödinger equation have the same energy E . These states are said to be degenerate. For example, the free particle states are doubly degenerate-one solution representing motion to the right. And the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension45 there are no degenerate bound states. Hint: Suppose there are two solutions, ψ1and ψ2with the same energy E. Multiply the Schrödinger equation for ψ1by ψ2and the Schrödinger equation for ψ2by ψ1and subtract, to show that ψ2dψ1/dx-ψ2dψ1/dxis a constant. Use the fact that for normalizable solutions ψ→0at±∞to demonstrate that this constant is in fact zero.Conclude that ψ2s a multiple of ψ1and hence that the two solutions are not distinct.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.