定积分的计算（分部积分法）

前置知识

定积分分部积分法

设 u , v u,v u,v在 [ a , b ] [a,b] [a,b]上可导，且 u ′ , v ′ u',v' u′,v′在 [ a , b ] [a,b] [a,b]上连续，则

∫ a b u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) ∣ a b − ∫ a b u ′ ( x ) v ( x ) d x \int_a^bu(x)v'(x)dx=u(x)v(x)\bigg\vert_a^b-\int_a^bu'(x)v(x)dx ∫ab​u(x)v′(x)dx=u(x)v(x) ​ab​−∫ab​u′(x)v(x)dx

也可写作

∫ a b u d v = u v ∣ a b − ∫ a b v d u \int_a^budv=uv\bigg\vert_a^b-\int_a^bvdu ∫ab​udv=uv ​ab​−∫ab​vdu

证明：根据复合函数求导，有

u ( x ) v ( x ) ∣ a b = ∫ a b [ u ( x ) v ( x ) ] ′ d x = ∫ a b u ( x ) v ′ ( x ) d x + ∫ a b u ′ ( x ) v ( x ) d x u(x)v(x)\bigg\vert_a^b=\int_a^b[u(x)v(x)]'dx=\int_a^bu(x)v'(x)dx+\int_a^bu'(x)v(x)dx u(x)v(x) ​ab​=∫ab​[u(x)v(x)]′dx=∫ab​u(x)v′(x)dx+∫ab​u′(x)v(x)dx

移项得

∫ a b u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) ∣ a b − ∫ a b u ′ ( x ) v ( x ) d x \int_a^bu(x)v'(x)dx=u(x)v(x)\bigg\vert_a^b-\int_a^bu'(x)v(x)dx ∫ab​u(x)v′(x)dx=u(x)v(x) ​ab​−∫ab​u′(x)v(x)dx

例题1

计算 ∫ 0 1 x e x d x \int_0^1xe^xdx ∫01​xexdx

解：

\qquad 原式 = ∫ 0 1 x d ( e x ) = x e x ∣ 0 1 − ∫ 0 1 e x d x = x e x ∣ 0 1 − e x ∣ 0 1 = e − ( e − 1 ) = 1 =\int_0^1xd(e^x)=xe^x\bigg\vert_0^1-\int_0^1e^xdx=xe^x\bigg\vert_0^1-e^x\bigg\vert_0^1=e-(e-1)=1 =∫01​xd(ex)=xex ​01​−∫01​exdx=xex ​01​−ex ​01​=e−(e−1)=1

例题2

已知 f ( 1 ) = 2 f(1)=2 f(1)=2， ∫ 0 1 f ( x ) d x = 1 \int_0^1f(x)dx=1 ∫01​f(x)dx=1，计算 ∫ 0 1 x f ′ ( x ) d x \int_0^1xf'(x)dx ∫01​xf′(x)dx

解：

\qquad 原式 = ∫ 0 1 x d [ f ( x ) ] = x f ( x ) ∣ 0 1 − ∫ 0 1 f ( x ) d x = f ( 1 ) − ∫ 0 1 f ( x ) d x = 2 − 1 = 1 =\int_0^1xd[f(x)]=xf(x)\bigg\vert_0^1-\int_0^1f(x)dx=f(1)-\int_0^1f(x)dx=2-1=1 =∫01​xd[f(x)]=xf(x) ​01​−∫01​f(x)dx=f(1)−∫01​f(x)dx=2−1=1