Prerequisite knowledge: integration by parts
Fractional integral
If fff in[ a , b ] [a,b][a,b ] is continuous, andc ∈ ( a , b ) c\in(a,b)c∈(a,b ) , then there are
∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx
Sometimes, ccc is not necessarily in( a , b ) (a,b)(a,b ) , but the equation still holds.
Segmental integration is to divide the integrand function into several segments, integrate them separately, and then sum them up.
Example 1
calculate∫ 0 2 π ∣ sin x ∣ dx \int_0^{2\pi}|\sin x|dx∫02 p.m∣sinx∣dx
Solution:
\qquad原式= ∫ 0 π sin xdx + ∫ π 2 π − sin xdx = − cos x ∣ 0 π + cos x ∣ π 2 π = 4 =\int_0^{\pi}\sin xdx+\int_{ \pi}^{2\pi}-\sin xdx=-\cos x\bigg\vert_0^{\pi}+\cos x\bigg\vert_{\pi}^{2\pi}=4=∫0psinx d x+∫Pi2 p.m−sinx d x=−cosx
0p+cosx
Pi2 p.m=4
Example 2
已知 f ( x ) = { e x , x ≤ 0 2 x + 2 , x > 0 f(x)=\begin{cases} e^x,\quad x\leq 0\\ 2x+2, \quad x>0 \end{cases} f(x)={ ex,x≤02x _+2,x>0，求 ∫ − 1 1 f ( x ) d x \int_{-1}^1f(x)dx ∫−11f(x)dx
Solution:
\qquad原式 = ∫ 0 1 e x d x + ∫ − 1 0 ( 2 x + 2 ) d x = e x ∣ 0 1 + ( x 2 + 2 x ) ∣ − 1 0 = e − 1 + 1 = e =\int_0^1e^xdx+\int_{-1}^0(2x+2)dx=e^x\bigg\vert_0^1+(x^2+2x)\bigg\vert_{-1}^0=e-1+1=e =∫01ex d x+∫−10( 2x _+2)dx=ex
01+(x2+2 x )
−10=e−1+1=e