Harman Outline 1A1 Integral Calculus CENG 5131

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1 Hrmn Outline 1A1 Integrl Clculus CENG 5131 September 5, 213 III. Review of Integrtion A.Bsic Definitions Hrmn Ch14,P642 Fundmentl Theorem of Clculus The fundmentl theorem of clculus shows the intimte reltionship between the derivtive nd the integrl. This theorem llows the evlution of integrls without computing f(x) from its definition s the limit of sum. To pply the theorem, the ntiderivtive of f(x) is defined s function F (x) such tht F (x) = f(x). Fundmentl Theorem of Clculus If f(x) is continuous function in the closed intervl [, b], then f(x) = F (b) F (), where F is n ntiderivtive of f. Assuming the function f(x) exists, this theorem implies tht df (x) = f(x) x b. Integrtion by Prts Consider the formul for the derivtive of product of functions, d [f(x)g(x)] = f(x)g (x) + g(x)f (x). Integrting both sides of the product eqution yields f(x)g(x) = f(x)g (x) + g(x)f (x), or f(x)g (x) = f(x)g(x) g(x)f (x). (1) This is the formul for integrtion by prts. It is usully written in condensed form by letting u = f(x), v = g(x), du = f (x), dv = g (x), so tht the formul for integrtion by prts becomes u dv = uv v du. (2) 1

2 y f(x+ x) f(x) x ε x dy y x x+ x x Figure 1: Cption for F12 4 new Consider the rc length in the figure. The length of smll segment is ] ds 2 = () 2 + (dy) 2 = [1 + (dy)2 () 2 () 2 Thus, the length of the curve is the sum of the smll line segment lengths. If the curve is continuous, the integrl of ds over the length in x is s = 1 + y 2 2

3 The integrl is used in vrious wys in every field of science nd engineering. Some exmples re given in the tble ssuming tht the function f(x) is continuous in the intervl considered. (Hrmn P. 649) Function Integrl Nme Are defined by y = f(x) A = f(x), f(x) Are 1 Averge of f(x) f = f(x) Averge on [, b] b Length of y = f(x) s = 1 + y 2 Arc Length Tble 1: Tble of Integrtion EXAMPLE of Integrtion by Prts To evlute the integrl π x sin x, we pply the integrtion by prts formul of Eqution 2. There re two possible choices for the vribles s u = x, dv = sin x or u = sin x, dv = x. The first choice simplifies the integrl to be evluted by prts, with the result π x sin x = x cos x π π ( cos x) = π cos π + cos + sin π sin = π. INT Integrte (Symbolic Mtlb) INT(S,v) is the indefinite integrl of S with respect to v. v is sclr SYM. INT(S,,b) is the definite integrl of S with respect to its symbolic vrible from to b. nd b re ech double or symbolic sclrs. INT(S,v,,b) is the definite integrl of S with respect to v from to b. Exmple: >> syms x >> int(x*sin(x)) ns = sin(x) - x*cos(x) >> int(x*sin(x),,pi) ns =pi (As Expected) 3

4 Prehps the most importnt integrl opertions in mthemticl physics nd engineering re the Fourier Trnsform, Lplce trnsform, nd the opertion of convolution Function Integrl Nme Fourier F (ω) = Lplce F (s) = Convolution y(t) = f(t)e iωt dt f(t)e ist dt f(τ)h(t τ)dτ ω = 2πf rd/sec s = α + iω Convolve f(t) nd h(t) Tble 2: Tble of Trnsforms nd Opertions See Hrmn P396 Ch8, P417 Ch9, nd P444. Notice tht integrtion by prts will often be used to compute the Fourier nd Lplce trnsforms. Convolution requires proper selection of the intervl of integrtion. EXAMPLE of Convolution Using the definition of the Unit Step Function, {, t <, U(t) = 1, t. we wish to convolve h(t) = e αt U(t+3) nd f(t) = e αt U(t 1). The importnt thing here is to recognize tht x(t) is zero before t = 3 nd h(t) is zero until t = 1. However, in the convolution formul f(τ) = e ατ U(τ 1) s before but h(t τ) = e αt τ U(t τ + 3) The convolution integrl becomes y(t) = [e ατ U(τ 1)][e α(t τ) U(t τ + 3)]dτ Since e αt is not function of τ, we cn write y(t) = e αt [U(τ 1)][U(t τ + 3)]dτ The lower limit is t = 1 nd the upper is t τ + 3 = or τ = t + 3 nd thus t+3 y(t) = e αt dτ = e αt [t + 3 1] = (t + 2)e αt U(t + 2) 1 where the restriction of the rnge of y(t) comes from the limits t or t 2. 4

5 3. EXAMPLE Physicl Applictions of Integrls In physics nd engineering, derivtives (rte of chnge) nd integrls define mny importnt properties of physicl systems. Derivtive forms of physicl reltionships re given in the tble. Function Derivtive Nme Electricl E field Force L di dt, C dv dt, i = dq dt E = dv F = d(mv) dt L nd C constnt Electric field strength (volts/m) is rte of chnge of voltge with distnce Newton s 2nd Lw Power P = dw dt If power in kilowtts, work W in kilowtt-hrs Conservtive force Het Conduction F (x) = du(x) dq dt = kadt Force equls chnge in potentil energy Het flow cross re A due to temperture grdient k is therml conductivity Tble 3: Tble of Physicl Reltionships Some comments on the entries in the tble re s follows: 1. The electricl equtions re ccurte when the vlues L nd C re constnt. The E field is vector in generl. 2. The net force on prticle is equl to the time rte of chnge of its liner momentum: F = d(mv)/dt. If m is constnt then the fmilir form F = m is used where F is vector in generl. 3. The work done on body by force is equl to the chnge in kinetic energy of the body. Power is the rte t which work is done. 5

6 3. EXAMPLE Physicl Applictions of Integrls In physics nd engineering, the integrl form of equtions is used when summtion of vrible is needed. Function Reltionship Nme velocity v(t) = Voltge V = (t) dt E(x) (t) is ccelertion Work nd Force W = Het (Btu) Q = m Gs Volume V W = t pressure P F (x) TF Vf Work moving object from to b c(t ) dt Het to mss m to increse T i temperture from T i to T f c is specific het V i P dv Work from expnsion Tble 4: Tble of Physicl Reltionships Some comments on the entries in the tble re s follows: 1. Velocity nd ccelertion nd the electric E field re vectors generlly. 2. If the forces or fields re conservtive mening no loss of energy if n object is moved through closed pth, the work is sid to be independent of pth. 6