Course 2BA1 Supplement concerning Integrtion by Prts Dvi R. Wilkins Copyright c Dvi R. Wilkins 22 3 The Rule for Integrtion by Prts Let u n v be continuously ifferentible rel-vlue functions on the intervl [, b]. Then u v = [uv]b v u, where [uv] b = u(b)v(b) u()v(). (A function is si to be continuously ifferentible if it is ifferentible n its erivtive is continuous function.) Derivtion of the Rule for Integrtion by Prts The rule for Integrtion by Prts is consequence of the Funmentl Theorem of Clculus n the Prouct Rule for Differentition. We begin with the Funmentl Theorem of Clculus. Funmentl Theorem of Clculus. Let f be continuous rel-vlue function on the intervl [, b]. Then x f(t) t = f(x) for ll x stisfying < x < b. Another wy of expressing the Funmentl Theorem of Clculus is the following: ( ) f(x) + C = f(x), 1
where f(x) is n inefinite integrl of the function f n C is n rbitrry rel constnt (often referre to s constnt of integrtion ). From the Funmentl Theorem of Clculus we cn erive the following bsic result: Corollry. Let f be continuously ifferentible rel-vlue function on the intervl [, b]. Then = f(b) f(). In orer to erive this result from the Funmentl Theorem of Clculus, efine function G on the intervl [, b] s follows: Then G() =, n G(s) = f(s) f() f(s) G(s) = s s s s s. f(s) = s f(s) s for ll rel numbers s stisfying < s < b. The fct tht the erivtive of the function G is zero everywhere in the interior of the intervl [, b] is sufficient to ensure tht the function is constnt on the intervl, n thus = G() = G(b) = f(b) f() n this ientity my be rerrnge to yiel the require result. In orer to erive the rule for Integrtion by Prts, we pply the corollry erive bove in the cse where the function f is the prouct of two continuously ifferentible rel-vlue functions u n v. We fin tht, = (u(x)v(x)) = [u(x)v(x)]b = u(b)v(b) u()v(). But the Prouct Rule for ifferentition yiels (u(x)v(x)) = u(x)v(x) + v(x). 2
Therefore n thus u(x) v(x) + u(x) v(x) v(x) = u(b)v(b) u()v() which is the rule for Integrtion by Prts. = u(b)v(b) u()v(), v(x), Exmples of the Use of the Rule for Integrtion by Prts Exmple We evlute the integrl Prts. We efine u n v so tht We my therefore tke u(x) = x, 1 x(x + 2) 3 using Integrtion by v(x) = (x + 2)3. v(x) = 1 (x + 2), n note tht = 1. On substituting in these functions into the formul for Integrtion by Prts, we fin tht 1 [ ] 1 1 x(x + 2) 3 = x(x + 2) 1 1 (x + 2) = 1 ( 1 3 1 ) 1 [ ] (x + 2) 5 1 2 = 81 23 32 5 211 = = 19 2 2 2 = 97 1. Exmple We evlute the integrl We tke u(x) = x, v(x) = e sx, = 1, v(x) = 1 s e sx, 3 xe sx using Integrtion by Prts.
n pply the rule for Integrtion by Prts, which yiels [ xe sx = 1 ] c s xe sx ( 1s ) e sx = c s e sc + 1 s e sx = c s e sc + 1 s 2 (1 e sc ) If s >, we my tke the limit s c, which gives us xe sx = 1 s 2. Exmple One cn use the Principle of Mthemticl Inuction n the metho of Integrtion by Prts to evlute x n e sx for ll positive integers n n positive rel numbers s. Let n be positive integer, n let s > n c >. We tke u(x) = x n, v(x) = e sx, = nxn 1, v(x) = 1 s e sx, n pply the rule for Integrtion by Prts, which yiels [ x n e sx = 1 ] c s xn e sx ( n ) s xn 1 e sx = cn s e sc + n s x n 1 e sx Now if s > then lim x n e sx. Thus if we tke the limit s c, we x fin tht x n e sx = n s A simple proof by inuction on n now shows tht x n e sx = n! s n+1 for ll positive integers n, provie tht s >. x n 1 e sx.
Exmple We evlute cos n x. Let us choose the functions u n v so tht u(x) = cos n 1 v(x) x, = cos x, = (n 1) cosn 2 x sin x, v(x) = sin x. Using the rule for Integrtion by Prts, we fin tht cos n x = cos n 1 x cos x = [ cos n 1 x sin x ] π = (n 1) cos n 2 x sin 2 x ( (n 1) cos n 2 x sin 2 x ) since sin x = when x = n when x = π. But sin 2 x = 1 cos 2 x. Therefore cos n x = (n 1) cos n 2 x (n 1) cos n x. On rerrnging this ientity, we fin tht for ll nturl numbers n. Now It follows tht cos n x = n 1 n cos n 2 x cos x = [sin x] π =. cos n x = for ll o positive integers n. When n is n even positive integer, we my set n = 2m for some positive integer m. A strightforwr proof by inuction on m then shows tht for ll positive integers m. cos 2m x = (2m)! π 2 2m (m!) 2 5