Prerequisite knowledge: Calculation of definite integrals (segmented integrals)
Exercise 1
Calculate ∫ 1 ee ∣ ln x ∣ dx \int_{\frac 1e}^e|\ln x|dx∫e1e∣lnx∣dx
Solution:
\qquad原式 = ∫ 1 e 1 − ln x d x + ∫ 1 e ln x d x = − ln x ∣ 1 e 1 + ln x ∣ 1 e = 1 + 1 = 2 =\int_{\frac 1e}^1-\ln xdx+\int_1^e\ln xdx=-\ln x\bigg\vert_{\frac 1e}^1+\ln x\bigg\vert_1^e=1+1=2 =∫e11−lnx d x+∫1elnx d x=−lnx
e11+lnx
1e=1+1=2
Exercise 2
已知 f ( x ) = { 1 1 + x 2 , x ≥ 0 x e x 2 , x < 0 f(x)=\begin{cases} \dfrac{1}{1+x^2},\quad x\geq 0\\ \qquad \\ xe^{x^2},\quad x<0 \end{cases} f(x)=⎩ ⎨ ⎧1+x21,x≥0x ex2,x<0, calculate ∫ − 1 1 f ( x ) dx \int_{-1}^1f(x)dx∫−11f(x)dx
Solution:
\qquad原式 = ∫ − 1 0 x e x 2 d x + ∫ 0 1 1 1 + x 2 d x =\int_{-1}^0xe^{x^2}dx+\int_0^1\dfrac{1}{1+x^2}dx =∫−10x ex2dx+∫011+x21dx
= 1 2 e x 2 ∣ − 1 0 + arctan x ∣ 0 1 = 1 2 − e 2 + π 4 \qquad\qquad =\dfrac 12e^{x^2}\bigg\vert_{-1}^0+\arctan x\bigg\vert_0^1=\dfrac 12-\dfrac{e}{2}+\dfrac{\pi}{4} =21ex2 −10+arctanx 01=21−2e+4p
Exercise 3
已知 f ( x ) = { 1 + x 2 , x ≤ 0 e − x , x > 0 f(x)=\begin{cases} 1+x^2,\quad x\leq 0\\ e^{-x},\quad x>0 \end{cases} f(x)={ 1+x2,x≤0e−x,x>0, calculate ∫ 1 3 f ( x − 2 ) dx \int_1^3f(x-2)dx∫13f(x−2)dx
Solution:
\qquad原式 = ∫ 1 2 [ 1 + ( x − 2 ) 2 ] d x + ∫ 2 3 e 2 − x d x = ∫ 1 2 ( x 2 − 4 x + 5 ) d x + ∫ 2 3 e 2 − x d x =\int_1^2[1+(x-2)^2]dx+\int_2^3e^{2-x}dx=\int_1^2(x^2-4x+5)dx+\int_2^3e^{2-x}dx =∫12[1+(x−2)2]dx+∫23e2 − x dx=∫12(x2−4x _+5)dx+∫23e2 − x dx
= ( 1 3 x 3 − 2 x 2 + 5 x ) ∣ 1 2 − e 2 − x ∣ 2 3 = 14 3 − 10 3 − 1 e + 1 = 7 3 − 1 e \qquad\qquad =(\dfrac 13x^3-2x^2+5x)\bigg\vert_1^2-e^{2-x}\bigg\vert_2^3=\dfrac{14}{3}-\dfrac{10}{3}-\dfrac 1e+1=\dfrac 73-\dfrac1e =(31x3−2x _2+5 x ) 12−e2−x 23=314−310−e1+1=37−e1