∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^bf(x)dx=\int_a^bf(a+b-x)dx ∫abf(x)dx=∫abf(a+b−x)dx
证明采用换元法即可,略。
[例1] 求: ∫ 0 1 x ( 1 − x ) 3 d x \int_0^1x(1-x)^3dx ∫01x(1−x)3dx
解:
∫ 0 1 x ( 1 − x ) 3 d x ⟹ 区间再现 ∫ 0 1 x 3 ( 1 − x ) d x = 1 20 \int_0^1x(1-x)^3dx\ \stackrel{区间再现}{\Longrightarrow}\int_0^1x^3(1-x)dx=\frac 1 {20} ∫01x(1−x)3dx ⟹区间再现∫01x3(1−x)dx=201
∫ 0 π 2 f ( sin x ) d x = ∫ 0 π 2 f ( cos x ) d x \int_0^{\frac \pi 2}f(\sin x)dx=\int_0^{\frac \pi 2}f(\cos x)dx ∫02πf(sinx)dx=∫02πf(cosx)dx
[例2] 求: ∫ 0 π 2 sin 2 x d x \int_0^{\frac \pi 2}\sin^2xdx ∫02πsin2xdx
解:
∫ 0 π 2 sin 2 x d x = ∫ 0 π 2 cos 2 x d x = 1 2 ∫ 0 π 2 d x = π 4 \int_0^{\frac \pi 2}\sin^2xdx=\int_0^{\frac \pi 2}\cos^2xdx=\frac 1 2\int_0^{\frac \pi 2}dx=\frac \pi 4 ∫02πsin2xdx=∫02πcos2xdx=21∫02πdx=4π
∫ 0 π x f ( sin x ) d x = π 2 ∫ 0 π f ( sin x ) d x \int_0^\pi xf(\sin x)dx=\frac \pi 2\int_0^\pi f(\sin x)dx ∫0πxf(sinx)dx=2π∫0πf(sinx)dx
[例3] 求: ∫ 0 π x sin 2 x d x \int _0^\pi x\sin ^2xdx ∫0πxsin2xdx
解:
∫ 0 π x sin 2 x d x = π 2 ∫ 0 π sin 2 x d x = π ∫ 0 π 2 sin 2 x d x = π 2 4 \int _0^\pi x\sin ^2xdx=\frac \pi 2\int_0^\pi \sin^2xdx=\pi\int_0^{\frac \pi 2}\sin^2xdx=\frac{\pi^2}{4} ∫0πxsin2xdx=2π∫0πsin2xdx=π∫02πsin2xdx=4π2
证明均采用区间再现,略。
∫ 0 T x f ( x ) d x = T 2 ∫ 0 T f ( x ) d x ,其中 f ( x ) = f ( 2 T − x ) \int_0^Txf(x)dx=\frac{T}{2}\int_0^Tf(x)dx,其中f(x)=f(2T-x) ∫0Txf(x)dx=2T∫0Tf(x)dx,其中f(x)=f(2T−x)
[例4] 求: ∫ 0 n π x ∣ sin x ∣ d x \int _0^{n\pi}x|\sin x|dx ∫0nπx∣sinx∣dx
解:考虑到, ∣ sin x ∣ = ∣ sin ( n π − x ) ∣ |\sin x|=|\sin(n\pi-x)| ∣sinx∣=∣sin(nπ−x)∣
于是:
∫ 0 n π x ∣ sin x ∣ d x = n π 2 ∫ 0 n π ∣ sin x ∣ d x = n 2 π 2 ∫ 0 π sin x d x = n 2 π \int _0^{n\pi}x|\sin x|dx=\frac{n\pi}{2}\int_0^{n\pi}|\sin x|dx=\frac{n^2\pi}{2}\int_0^{\pi}\sin xdx=n^2\pi ∫0nπx∣sinx∣dx=2nπ∫0nπ∣sinx∣dx=2n2π∫0πsinxdx=n2π
∫ a b f ( x ) f ( x ) + f ( a + b − x ) d x = b − a 2 \int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2} ∫abf(x)+f(a+b−x)f(x)dx=2b−a
[例5] 求: ∫ 2 4 x x + 6 − x d x \int _2^4\frac{\sqrt x}{\sqrt x+\sqrt{6-x}}dx ∫24x+6−xxdx
解:
∫ 2 4 x x + 6 − x d x = 4 − 2 2 = 1 \int _2^4\frac{\sqrt x}{\sqrt x+\sqrt{6-x}}dx=\frac{4-2}{2}=1 ∫24x+6−xxdx=24−2=1
∫ 0 π 2 f ( sin x ) f ( sin x ) + f ( cos x ) d x = ∫ 0 π 2 f ( cos x ) f ( sin x ) + f ( cos x ) d x = π 4 \int_0^{\frac \pi 2}\frac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int_0^{\frac \pi 2}\frac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\frac{\pi}{4} ∫02πf(sinx)+f(cosx)f(sinx)dx=∫02πf(sinx)+f(cosx)f(cosx)dx=4π
[例6]:求:
∫ 0 π 2 e sin x e sin x + e cos x d x \int_0^{\frac \pi 2} \frac{e^{\sin x}}{e^{\sin x}+e^{\cos x}}dx ∫02πesinx+ecosxesinxdx
解:
∫ 0 π 2 e sin x e sin x + e cos x d x = π 4 \int_0^{\frac \pi 2} \frac{e^{\sin x}}{e^{\sin x}+e^{\cos x}}dx=\frac{\pi}{4} ∫02πesinx+ecosxesinxdx=4π
若 f ( x ) f(x) f(x)在 [ − a , a ] [-a,a] [−a,a]连续,则:
∫ − a a f ( x ) d x = ∫ 0 a [ f ( x ) + f ( − x ) ] d x \int_{-a}^af(x)dx=\int_0^a[f(x)+f(-x)]dx ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx
[例7]:求:
∫ − π 2 π 2 sin 2 x 1 + e x d x \int_{-\frac{\pi}{2}}^{\frac \pi 2}\frac{\sin^2 x}{1+e^{x}}dx ∫−2π2π1+exsin2xdx
解:
∫ − π 2 π 2 sin 2 x 1 + e x d x = ∫ 0 π 2 ( sin 2 x 1 + e x + sin 2 x 1 + e − x ) d x = ∫ 0 π 2 sin 2 x d x = π 4 \int_{-\frac{\pi}{2}}^{\frac \pi 2}\frac{\sin^2 x}{1+e^{x}}dx=\int_{0}^{\frac \pi 2}(\frac{\sin^2 x}{1+e^{x}}+\frac{\sin^2 x}{1+e^{-x}})dx=\int_0^{\frac \pi 2}\sin^2xdx=\frac{\pi}{4} ∫−2π2π1+exsin2xdx=∫02π(1+exsin2x+1+e−xsin2x)dx=∫02πsin2xdx=4π