⑩天掌 Teaching Plan on Advanced Mathematics 第三节定积分的换元法和分部积分法 定积分的换元法 、定积分的分部积分法 「返回 Tianjin Polytechnic Moiwendity w
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 第三节 定积分的换元法和分部积分法 一、定积分的换元法 二、定积分的分部积分法 返回
⑩天掌 Teaching Plan on Advanced Mathematics 定积分的换元法 定理假设 (1)f(x)在[a,b上连续; (2)函数x=φ()在[a,上是单值的且有连续导数; (3)当t在区间a,B上变化时,x=q(t)的值在a,b]上 变化,且φp(a)=a、p(B)=b, 则有f(x)x=np()lp()t
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 定理 假设 （1） f (x)在[a,b]上连续； （2）函数x = (t)在[, ]上是单值的且有连续导数； （3）当t在区间[, ]上变化时， x = (t)的值在[a,b] 上 变化，且() = a、( ) = b， 则 有 f x dx f t t dt b a = ( ) [( )] ( ) . 一、定积分的换元法
⑩天掌 Teaching Plan on Advanced Mathematics qp(a)=a、q(B)=b, ⑩()-(a)=F|q(B)-F|p(a) =F(b)-F(a) f(x=F(b)-(a)=0)-(a) IfIp(tlo(t)dt. 注意当a>B时,换元公式仍成立
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics () = a、( ) = b, ( ) − () = F[( )]− F[()] = F(b) − F(a), f (x)dx F(b) F(a) b a = − = ( ) − () f [ (t)] (t)dt. = 注意 当 时，换元公式仍成立
⑩天掌 Teaching Plan on Advanced Mathematics 应用换元公式时应注意: (1)用x=g(t)把变量x换成新变量t时,积分限也 相应的改变 (2)求出∫|p()p(t)的一个原函数Φ()后,不必象 计算不定积分那样再要把Φ(t)变换成原变量x 的函数,而只要把新变量t的上、下限分别代入 Φ(t)然后相减就行了
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 应用换元公式时应注意: 求出 f[(t)](t)的一个原函数(t)后，不必象 计算不定积分那样再要把(t)变换成原变量x 的函数，而只要把新变量t的上、下限分别代入 (t)然后相减就行了. （2） （1） 用x = (t)把变量x换成新变量t时，积分限也 相应的改变
⑩天掌 Teaching Plan on Advanced Mathematics 例1计算 20 cos r sin xe dx 解令t=c0sx,t=- sinxdx, 2 →t=0,x=0→t=1. cosxsin xdx tdt 66
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例1 计算 cos sin . 2 0 5 x xdx 令 2 x = t = 0, x = 0 t = 1, 2 0 5 cos x sin xdx = − 0 1 5 t dt 1 0 6 6 t = . 6 1 = 解 t = cos x, dt = −sin xdx
⑩天掌 Teaching Plan on Advanced Mathematics 例2计算 sin'x-sinxdx 3 解:∫(x)=√sin3x-sin°x=cosx(sinx) T sin'x-sin xdx=l cos x(sin x 2dx 2 cos x(sinx2dx cosrIsinr )2d 2 sinx)id sinx-[(sin sin SIn x SIn 0 5
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例2 计算 sin sin . 0 3 5 x − xdx 解 f x x x 3 5 ( ) = sin − sin ( )2 3 = cos x sin x − 0 3 5 sin x sin xdx ( ) = 0 2 3 cos x sin x dx ( ) = 2 0 2 3 cos x sin x dx ( ) − 2 2 3 cos x sin x dx ( ) = 2 0 2 3 sin x d sin x ( ) − 2 2 3 sin x d sin x ( ) 2 0 2 5 sin 5 2 = x ( ) − 2 2 5 sin 5 2 x . 5 4 =
⑩天掌 Teaching Plan on Advanced Mathematics d x 例3计算 ex√lnx(1-lnx) 解原式 d(Inx) nx nx d(ln x) d √Inx n nx 2 nx T 2arcsin(Inx
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例3 计算 . ln (1 ln ) 4 3 − e e x x x dx 解 原式 − = 4 3 ln (1 ln ) e (ln ) e x x d x − = 4 3 ln (1 ln ) e (ln ) e x x d x − = 4 3 2 1 ( ln ) ln 2 e e x d x 4 3 2 arcsin( ln ) e e = x . 6 =
⑩天掌 Teaching Plan on Advanced Mathematics 例4计算 dx.(a>0) x+√a-x 解令x= aint,dx= acos tdi, x=a→t= x=0→t=0 2 acos t 原式 d t aint +a(1-sinft cos t cos t- sin t 十 0 sint+ cos t sin t+ cos t -+-|nsnt+cos/2、x 0
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例4 计算 + − a dx a x a x 0 2 2 . ( 0) 1 x = a , 2 t = x = 0 t = 0, 解 令 x = asint, dx = acostdt, 原式 + − = 2 0 2 2 sin (1 sin ) cos dt a t a t a t + = 2 0 sin cos cos dt t t t + − = + 2 0 sin cos cos sin 1 2 1 dt t t t t 2 0 lnsin cos 2 1 2 2 1 + + = t t . 4 =
⑩天掌 Teaching Plan on Advanced Mathematics 例5当f(x)在-a,a上连续,且有 ①f(x)为偶函数,则 f(x)dx=2 f(x)dx ②f(x)为奇函数,则f(x)t=0 证 f(x)=」f(x)d+f(x)d, 在。f(x)中令x=t
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例 5 当 f (x)在[−a, a]上连续，且有 ① f (x)为偶函数，则 − = a a a f x dx f x dx 0 ( ) 2 ( ) ； ② f (x)为奇函数，则− = a a f (x)dx 0. 证 ( ) ( ) ( ) , 0 0 − − = + a a a a f x dx f x dx f x dx 在− 0 ( ) a f x dx中令x = −t
⑩天掌 Teaching Plan on Advanced Mathematics ∫。f(xt=-(n)d=f(-l, ①f(x)为偶函数,则f(-t)=∫(t) ,(x)=f(x)d+(x =2|(d ②f(x)为奇函数,则f(-)=-f(t) f(x)d=.,f(x)+(=0
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics − = 0 ( ) a f x dx − − = 0 ( ) a f t dt ( ) , 0 − a f t dt ① f (x)为偶函数，则 f (−t) = f (t), − − = + a a a a f x dx f x dx f x dx 0 0 ( ) ( ) ( ) 2 ( ) ; 0 = a f t dt ② f (x)为奇函数，则 f (−t) = − f (t), − = − + a a a a f x dx f x dx f x dx 0 0 ( ) ( ) ( ) = 0