例3.设f(x)∈C[-a,a], 偶倍奇零 ().(-x)=f(),则。(x)dx=2」0/(x)dx 2)若f(-x)=-f(x),则f(x)dx=0 ME: f(x)dx=. f(x)dx+f(x)d: 「/()d7+0(x)d令x =0[f(-x)+f(x)dx 20/(x)dx,f(x)=f(x)时 0 f(-x)=-f(x)时 HIGH EDUCATION PRESS 机动目录上页下页返回结
例3. 证: (1) 若 − = a a a f x x f x x 0 则 ( )d 2 ( )d = − f x x a a ( )d (2) 若 ( )d = 0 − a a 则 f x x f x x a ( )d 0 − f x x a ( )d 0 + f t t a ( )d 0 = − f x x a ( )d 0 + f x f x x a [ ( ) ( )]d 0 = − + f (−x) = f (x)时 f (−x) = − f (x)时 偶倍奇零 机动 目录 上页 下页 返回 结束 令x = −t =
二、定积分的分部积分法 定理2.设v(x),v(x)∈C[a,b],则 b urv(rr=u(rv(r u(rvrdx 证:∵[(x)v(x)=l(x)v(x)+l(x)(x) 两端在[a,b]上积分 b rb uav(x )=u(x)v(x dx+ u(x)v'(x)dx C b u(x)v'(x)dx=u(x)v(x) u(Xvr)ax HIGH EDUCATION PRESS 机动目录上页下页返回结
二、定积分的分部积分法 定理2. ( ), ( ) [ , ], 1 设u x v x C a b 则 a b 证: [u(x)v(x)] = u (x)v(x) + u(x)v (x) u(x)v(x) a b u x v x x u x v x x b a b a = ( ) ( )d + ( ) ( )d = u(x)v(x) a b − b a u (x) v(x)dx 机动 目录 上页 下页 返回 结束 两端在[a,b]上积分