()+x)x=∑()x+∑/(5)△x lim∑L(5k)+()x4-lim∑f(5)x4+lim∑/(5Axk (T)→0k=1 (T)→0k=1 叮f()+f(x)k=Jf(x+∫ Th3.f(x)在[ab]上可积,则cf(x)在[ab]上也可积,且 cf(x)dx=cf(x)dx 证明:∑(5)xk=c∑/(5)x 推论:n个函数f;(x),f2(x)…fn(x),都在区间[ab]上可积 则它们的线性组合 cf;(x)+c2.f1(x)+…+cnf(x)在[ab]上也可积 ∫c1f(x)+c2f(x)+…+f,(x)k=c∫f(x)hx+c2Jf2(x Th4,fx)在[ab]上可积则fx)在[a1,b3]sab上可积 Th5.f(x)在[a,b]与[cb]上可积,则f(x)在[a,b]上可积 f(x)dx= f(r)dx+f(x)dx 推论1若f(x)在[AB]上可积,且ab,c是[A,B]上任意三点 则 f(x)dx=f(x)dx+/(x)dr 推论2若f(x)在区间[lk-1,4k](k=1,2n)上都可积,则fx)在[l,ln]上可积 且 f()dx= f()dx+ f(x)dx +.+f(x)dx= +  n k k f k f x x 1 1 2 [ ( ) ( )] ＝ =  n k k f k x 1 1 ( ) + =  n k k f k x 1 2 ( ) lim l(T )→0 = +  n k k f k f k x 1 1 2 [ ( ) ( )] = lim l(T )→0 =  n k k f k x 1 1 ( ) + lim l(T )→0 =  n k k f k x 1 2 ( ) 即  + b a [ f (x) f (x)]dx 1 2 ＝  b a f (x)dx 1 ＋  b a f (x)dx 2 Th3. f(x)在［a,b］上可积，则 c f(x)在［a,b］上也可积，且  b a cf (x)dx ＝c  b a f (x)dx 证明： =  n k k cf k x 1 ( ) ＝c =  n k k f k x 1 ( ) 推论：n 个函数 ( ) 1 f x ， ( ) 2 f x … f (x) n 都在区间［a,b］上可积 则它们的线性组合 1 c ( ) 1 f x ＋ 2 c ( ) 2 f x ＋….＋ n c f (x) n 在［a,b］上也可积．  + + + b a n x f x f x dx c f c ( ) ( )] 2 ( ) 1 [ 1 2  ＝ c1  b a f (x)dx 1 ＋ c2  b a f (x)dx 2 ＋…＋  b a n f (x)dx Th4 .，f(x)在［a,b］上可积则 f(x)在［ 1 1 a ,b ,］  [a,b]上可积． Th5. f(x)在［a,b］与［c,b］上可积,则 f(x)在［a,b］上可积．  b a f (x)dx ＝ +  c a f (x)dx  b c f (x)dx 推论１ 若 f(x)在［A,B］上可积，且 a,b,c 是［A,B］上任意三点 则  b a f (x)dx ＝ +  c a f (x)dx  b c f (x)dx 推论２ 若 f(x)在区间［ k k l ,l −1 ］(k=1,2,…n)上都可积，则 f(x)在[ n l ,l 0 ]上可积， 且  n l l f x dx 0 ( ) ＝  1 0 ( ) l l f x dx ＋  2 1 ( ) l l f x dx ＋…＋  − n n l l f x dx 1 1 ( )