定积分的性质

有极限的积分称为定积分。它有上限和下限。它表示为

$\int_{a}^{b}$ f(x) = F(b) - F(a)

关于定积分有很多性质。我们还将用证据一一讨论每个属性。

特性

属性一： $\int_{a}^{b}$ f(x) dx = $\int_{a}^{b}$ f(y) dy

证明：

$\int_{a}^{b}$ f(x) dx…….(1)

Suppose x = y

           dx = dy

Putting this in equation (1)

$\int_{a}^{b}$ f(y) dy

属性 2： $\int_{a}^{b}$ f(x) dx = – $\int_{b}^{a}$ f(x) dx

证明：

$\int_{a}^{b}$ f(x) dx = F(b) – F(a)……..(1)

$\int_{b}^{a}$f(x) dx = F(a) – F(b)………. (2)

From (1) and (2) 

We can derive $\int_{a}^{b}$ f(x) dx = –$\int_{b}^{a}$ f(x) dx

属性 3： $\int_{a}^{b}$ f(x) dx = $\int_{a}^{p}$ f(x) dx + $\int_{p}^{b}$ f(x) dx

证明：

$\int_{a}^{b}$ f(x) dx = F(b) – F(a) ………..(1)

$\int_{a}^{p}$ f(x) dx = F(p) – F(a) ………..(2)

$\int_{p}^{b}$ f(x) dx = F(b) – F(p) ………..(3)

From (2) and (3)

$\int_{a}^{p}$f(x) dx +$\int_{p}^{b}$f(x) dx = F(p) – F(a) + F(b) – F(p)

$\int_{a}^{p}$f(x) dx + $\int_{p}^{b}$f(x) dx = F(b) – F(a) = $\int_{a}^{b}$f(x) dx    

Hence, it is Proved.

属性 4.1： $\int_{a}^{b}$ f(x) dx = $\int_{a}^{b}$ f(a + b – x) dx

证明：

Suppose 

a + b – x = y…………(1)

-dx = dy 

From (1) you can see 

when x = a  

y = a + b – a

y = b

and when x = b

y = a + b – b

y = a

Replacing by these values he integration on right side becomes $-\int_{b}^{a}$ f(y)dy

From property 1 and property 2 you can say that 

$\int_{a}^{b}$ f(x) dx = $\int_{a}^{b}$f(a + b – x) dx  

属性 4.2：如果 a 的值为 0，则属性 4.1 可以写为

 $\int_{0}^{b}$ f(x) dx = $\int_{0}^{b}$f(b – x) dx

属性 5： $\int_{0}^{2a}$ f(x) dx = $\int_{0}^{a}$ f(x) dx + $\int_{0}^{a}$ f(2a – x) dx

证明：

We can write $\int_{0}^{2a}$ f(x) dx as

$\int_{0}^{2a}$ f(x) dx = $\int_{0}^{a}$f(x) dx + $\int_{a}^{2a}$f(x) dx  ………….. (1)

I = I₁ + I₂ 

(from property 3)

Suppose 2a – x = y

-dx = dy

Also when x = 0

y = 2a, and when x = a

y = 2a – a = a

So, $\int_{0}^{a}$ f(2a – x)dx  can be written as 

$-\int_{2a}^{a}$f(y) dy = I₂

Replacing equation (1) with he value of I₂ we get 

$\int_{0}^{2a}$ f(x) dx = $\int_{0}^{a}$f(x) dx + $\int_{0}^{a}$f(2a – x) dx

属性 6： $\int_{0}^{2a}$ f(x) dx = 2 $\int_{0}^{a}$ f(x) dx；如果 f(2a – x) = f(x)

= 0;如果 f(2a – x) = -f(x)

证明：

From property 5 we can write $\int_{0}^{2a}$ f(x) dx as

$\int_{0}^{2a}$ f(x) dx =$\int_{0}^{a}$f(x) dx + $\int_{0}^{a}$f(2a – x) dx  ………….(1)

Part  1: If f(2a – x) = f(x) 

Then equation (1) can be written as 

$\int_{0}^{2a}$f(x) dx =$\int_{0}^{a}$f(x) dx + $\int_{0}^{a}$f(x) dx

This can be further written as 

$\int_{0}^{2a}$ f(x) dx = 2 $\int_{0}^{a}$f(x) dx

Part  2: If f(2a – x) = -f(x)

Then equation (1) can be written as 

 $\int_{0}^{2a}$ f(x) dx=$\int_{0}^{a}$f(x) dx – $\int_{0}^{a}$f(x) dx

This can be further written as 

$\int_{0}^{2a}$ f(x) dx= 0

属性 7： $\int_{-a}^{a}$ f(x) dx = $\int_{0}^{a}$ f(x) dx；如果一个函数是偶数，即 f(-x) = f(x)

= 0;如果一个函数是奇数，即 f(-x) = -f(x)

证明：

From property 3 we can write 

$\int_{-a}^{a}$ f(x) dx as 

$\int_{-a}^{a}$ f(x) dx = $\int_{-a}^{0}$f(x) dx + $\int_{0}^{a}$f(x) dx  ………(1)

Suppose 

$\int_{-a}^{0}$ f(x) dx = I1 ……(2)

Now, assume x = -y

So, dx = -dy

And also when x = -a then

y= -(-a) = a

and when x = 0 then, y = 0

Putting these values in equation (2) we get

I₁ = $-\int_{a}^{0}$ f(-y)dy

Using property 2, I₁ can be written as 

I₁ = $\int_{0}^{a}$ f(-y)dy

and using property 1 I₁ can be written  as 

I₁ = $\int_{0}^{a}$ f(-x)dx

Putting value of I₁ in equation (1), we get

$\int_{-a}^{a}$ f(x) dx = $\int_{0}^{a}$f(-x) dx +$\int_{0}^{a}$f(x) dx   ……….(3)

Part 1: When f(-x) = f(x)

Then equation(3) becomes

$\int_{-a}^{a}$ f(x) dx = $\int_{0}^{a}$f(x) dx + $\int_{0}^{a}$f(x) dx

$\int_{-a}^{a}$ f(x) dx = 2$\int_{0}^{a}$f(x) dx 

Part  2: When f(-x) = -f(x)

Then equation 3 becomes

$\int_{-a}^{a}$ f(x) dx = –$\int_{0}^{a}$f(x) dx +$\int_{0}^{a}$f(x) d

$\int_{-a}^{a}$ f(x)dx = 0

例子

示例1：我 = $\int_{0}^{1}$ x(1 – x) ⁹⁹分

解决方案：

Using  property  4.2 he given question can be written as 

$\int_{0}^{1}$ (1 – x) [1 – (1 – x)]⁹⁹ dx

$\int_{0}^{1}$ (1 – x) [1 – 1 + x]⁹⁹ dx

$\int_{0}^{1}$(1 – x)x⁹⁹ dx

$\begin{bmatrix} \frac{x^{100}}{100} - \frac{x^{101}}{101} \end{bmatrix}_{0}^{1}$

= 1/100 – 1/101

= 1 / 10100

示例 2 ：我 = $\int_{\frac{-1}{2}}^{\frac{-1}{2}}$ cos(x) 日志$\begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}$

解决方案：

f(x) = cos(x) log $\begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}$

 f(-x) = cos(-x) log $\begin{vmatrix} \frac{1-x}{1+x} \end{vmatrix}$

 f(-x) = -cos(x) log $\begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}$

f(-x) = -f(x)

Hence the function is odd. So, Using property 

$\int_{-a}^{a}$ f(x)dx = 0; if a function is odd i.e. f(-x) = -f(x) 

$\int_{\frac{-1}{2}}^{\frac{-1}{2}}$ cos(x) log $\begin{vmatrix} \frac{1+x}{1-x} \end{vmatrix}$ = 0

示例 3：我 = $\int_{0}^{5}$ [x] dx

解决方案：

$\int_{0}^{1}$ 0 dx + $\int_{1}^{2}$1 dx + $\int_{2}^{3}$ 2 dx +$\int_{3}^{4}$ 3 dx + $\int_{4}^{5}$ 4 dx  [using Property 3]

= 0 + [x]²₁ + 2[x]³₂  + 3[x]⁴₃ + 4[x]⁵₄ 

= 0 + (2 – 1) + 2(3 – 2) + 3(4 – 3) + 4(5 – 4)

= 0 + 1 + 2 + 3 + 4

= 10

示例 4：我 = $\int_{-1}^{2}$ |x| dx

解决方案：

$\int_{-1}^{0}$ (-x) dx + $\int_{0}^{2}$ (x) dx  [using Property 3] 

= -[x²/2]⁰_-1 + [x²/2]²₀  

= -[0/2 – 1/2] + [4/2 – 0]

= 1/2 + 2

= 5/2