微积分 | 积分性质——定义、推导与证明 第二部分

注：本文为 “微积分 | 积分性质” 相关合辑。
英文引文，机翻未校。
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csdn 篇幅所限，本文共两部分。

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微积分 | 积分性质——定义、推导与证明 第二部分

How to prove the integral properties?

如何证明积分性质？

Now that we’ve learned about the properties that can help us in simplifying indefinite and definite integrals, it’s time that we learn how some of these properties were derived. In this section, we’ll show you the proof for the addition, constant multiple, and reverse interval properties of integrals.
既然我们已经学习了有助于简化不定积分和定积分的性质，现在就来学习其中一些性质的推导过程。在本节中，我们将为你展示积分的加法性质、常数因子性质和区间反转性质的证明。

$$\begin{array}{rl}\int k\cdot f(x)dx& =k\int f(x)dx\\ \int [f(x)+g(x)]dx& =\int f(x)dx+\int g(x)dx\\ {\int }_a^{b}f(x)dx& =-{\int }_b^{a}f(x)dx\end{array}$$

We’ll need some of the properties and derivative rules we’ve learned from our differential calculus classes, so make sure to refresh your knowledge. In addition, recall that differentiation’s process is the inverse of integration.
我们需要用到一些从微分学课程中学到的性质和导数法则，因此请务必复习相关知识。此外，记住微分过程是积分的逆运算。

$$\begin{array}{rl}\frac{d}{dx}\int f(x)dx& =f(x)\\ \int f^{\prime }(x)dx& =f(x)+C\end{array}$$

Keep this relationship in mind when deriving the rest of the formulas discussed in the previous section. For now, let’s begin by deriving the constant multiple property of integrals.
在推导上一节讨论的其他公式时，请记住这种关系。现在，我们先从推导积分的常数因子性质开始。

Proving the constant multiple property of integrals

证明积分的常数因子性质

To prove this property, we’ll need to use the constant multiple property of derivatives,
为了证明这个性质，我们需要用到导数的常数因子性质
$\frac{d}{dx}k\cdot f(x)=k\frac{d}{dx}f(x)$.

What we’ll do here is take the derivative of $\int k\cdot f(x)dx$ and $k\int \cdot f(x)dx$ with respect to $x$.
我们要做的是对 $\int k\cdot f(x)dx$ 和 $k\int f(x)dx$ 分别求关于 $x$ 的导数。

$$\begin{array}{rl}\frac{d}{dx}\left[\int k\cdot f(x)dx\right]& =k\cdot f(x)\\ \frac{d}{dx}\left[k\cdot \int f(x)dx\right]& =k\cdot \frac{d}{dx}\int f(x)dx\\ & =k\cdot f(x)\\ & \\ \frac{d}{dx}\left[\int k\cdot f(x)dx\right]& =\frac{d}{dx}\left[k\cdot \int f(x)dx\right]\end{array}$$

For the two sides of the equation to be true, $\int k\cdot f(x)dx $ must be equal to , $k \int f(x)dx $.Hence,confirmingtheconstantmultiplepropertyofintegrals.为了使等式两边成立，$\int k \cdot f(x)dx$ 必须等于 $k\int f(x)dx$。因此，积分的常数因子性质得证。

Proving the sum property of integrals
证明积分的和性质

We’ll apply a similar process to prove the sum property of integrals using the sum rule for derivatives,
我们将采用类似的方法，利用导数的和法则来证明积分的和性质。

$\frac{d}{dx}[f(x)+g(x)]=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$.

Take the derivative of $\int [f(x)+g(x)]dx$ and $\int f(x) dx + \int f(x) dx $.
对 $\int [f(x)+g(x)]dx$ 和 $\int f(x)dx+\int g(x)dx$ 分别求导。

$$\begin{array}{rl}\frac{d}{dx}\int [f(x)+g(x)]dx& =f(x)+g(x)\\ \frac{d}{dx}\left[\int f(x)dx+\int g(x)dx\right]& =\frac{d}{dx}\int f(x)dx+\frac{d}{dx}\int g(x)dx\\ & =f(x)+g(x)\\ & \\ \frac{d}{dx}\int [f(x)+g(x)]dx& =\frac{d}{dx}\left[\int f(x)dx+\int g(x)dx\right]\end{array}$$

Since both sides of the equation are equal, the integral expressions will also be equal.
由于等式两边相等，因此积分表达式也相等。

$$\begin{array}{rl}\int [f(x)+g(x)]dx& =\int f(x)dx+\int g(x)dx\end{array}$$

This confirms the sum property of indefinite integrals. We can apply a similar process to confirm the sum property of definite integrals,
这就证明了不定积分的和性质。我们可以用类似的方法证明定积分的和性质：

${\int }_a^b[f(x)+g(x)]dx={\int }_a^{b}f(x)dx+\int g(x)dx$.

Proving the reverse length property of integrals

证明积分的区间反转性质

For this property, let’s recall the second part of the fundament theorem of calculus:
为了证明这个性质，让我们回顾微积分基本定理的第二部分：

${\int }_a^{b}f(x)dx=F(b)–F(a)$,
where $F(x)$ is the antiderivative of $f(x)$.
其中 $F(x)$ 是 $f(x)$ 的原函数。

We can use this theorem to rewrite $-{\int }_b^{a}f(x)dx$.
我们可以用这个定理来改写 $-{\int }_b^{a}f(x)dx$。

$$\begin{array}{rl}-{\int }_b^{a}f(x)dx& =-[F(a)–F(b)]\\ & =-F(a)+F(b)\\ & =F(b)–F(a)\\ & ={\int }_a^{b}f(x)dx\end{array}$$

Hence, we’ve shown that reversing the limits’ positions will simply change the resulting value of the definite integral.
因此，我们证明了交换积分上下限的位置只会改变定积分的结果的符号。

We can use similar processes when proving the rest of the properties. But we left that for you to work on your own and some, we have included in the exercises shown below!
我们可以用类似的方法证明其他性质。但我们把这部分留给你自己完成，其中一些性质我们已包含在下面的练习中！

Make sure to keep your notes handy though when working on the rest of the problems.
在做其余题目时，请务必准备好你的笔记。

Example 1
例 1

Use the fact that ${\int }_0^{6}f(x)dx=12$ and ${\int }_{-6}^{0}f(x)dx=20$ to determine the value of ${\int }_{-6}^{6}f(x)dx$.
已知 ${\int }_0^{6}f(x)dx=12$ 和 ${\int }_{-6}^{0}f(x)dx=20$，求 ${\int }_{-6}^{6}f(x)dx$ 的值。

Solution
解

We begin by recalling the definite integral property,
我们首先回顾定积分的性质：
${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$.

We can use this property to combine the two given definite integrals.
我们可以用这个性质来合并这两个给定的定积分。

$$\begin{array}{rl}{\int }_0^{6}f(x)dx+{\int }_{-6}^{0}f(x)dx& ={\int }_{-6}^{0}f(x)dx+{\int }_0^{6}f(x)dx\\ & ={\int }_{-6}^{6}f(x)dx\end{array}$$

We’ve shown that we can find the value of ${\int }_{-6}^{6}f(x)dx$ by adding the values of the definite integrals, ${\int }_0^{6}f(x)dx=12$ and ${\int }_{-6}^{0}f(x)dx=20$.
我们已经证明，可以通过将定积分 ${\int }_0^{6}f(x)dx=12$ 和 ${\int }_{-6}^{0}f(x)dx=20$ 的值相加来得到 ${\int }_{-6}^{6}f(x)dx$ 的值。

$$\begin{array}{rl}{\int }_{-6}^{6}f(x)dx& =12+20\\ & =32\end{array}$$

Hence, we have ${\int }_{-6}^{6}f(x)dx=32$.
因此，${\int }_{-6}^{6}f(x)dx=32$。

Example 2
例 2

Use the fact that ${\int }_0^{3}f(x)dx=18$ and ${\int }_0^{10}f(x)dx=40$ to determine the value of ${\int }_3^{10}f(x)dx$.
已知 ${\int }_0^{3}f(x)dx=18$ 和 ${\int }_0^{10}f(x)dx=40$，求 ${\int }_3^{10}f(x)dx$ 的值。

Solution
解

We’ll also be using the same property,
我们也将使用同样的性质
${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$,

to answer this problem.
来解决这个问题。

Rewrite the expression so that we can expression ${\int }_3^{10}f(x)dx$ in terms of ${\int }_0^{3}f(x)dx$ and ${\int }_0^{10}f(x)dx$.
改写表达式，以便用 ${\int }_0^{3}f(x)dx$ 和 ${\int }_0^{10}f(x)dx$ 来表示 ${\int }_3^{10}f(x)dx$。

$$\begin{array}{rl}{\int }_0^{10}f(x)dx& ={\int }_0^{3}f(x)dx+{\int }_3^{10}f(x)dx\end{array}$$

Substitute the values of the two definite integrals then isolate ${\int }_3^{10}f(x)dx$ on one side of the equation.
代入两个定积分的值，然后将 ${\int }_3^{10}f(x)dx$ 分离到等式的一边。
$$\begin{array}{rl}40& =18+{\int }_3^{10}f(x)dx\\ 40-18& ={\int }_3^{10}f(x)dx\\ {\int }_3^{10}f(x)dx& =22\end{array}$$
This means that by combining intervals, we were able to determine the value of ${\int }_3^{10}f(x)d$. In fact, it’s equal to $22$.
这意味着通过合并区间，我们可以确定 ${\int }_3^{10}f(x)dx$ 的值。实际上，它等于 $22$。

Example 3
例 3

Prove the definite integral property,
证明定积分性质：

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

Solution

解

Recall that when $F(x)$ is the antiderivative of $f(x)$, we have
回想一下，当 $F(x)$ 是 $f(x)$ 的原函数时，我们有

${\int }_a^{b}f(x)dx=F(b)–F(a)$.

Use the second part of the fundamental theorem of calculus to rewrite the right-hand side of the equation.
利用微积分基本定理的第二部分来重写等式的左边。

$$\begin{array}{rl}{\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx& =[F(b)–F(a)]+[F(c)–F(b)]\\ & =F(b)–F(a)+F(c)–F(b)\\ & =F(c)–F(a)\end{array}$$

Rewrite $F(c)–F(a)$ back into a definite integral.
将 $F(c)-F(a)$ 重新写成定积分的形式。

$$\begin{array}{rl}{\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx& =F(c)–F(a)\\ & ={\int }_a^{c}f(x)dx\end{array}$$

This confirms the definite integral property we use to combine intervals is indeed true.
这就证明了我们用于合并区间的定积分性质确实成立。

Example 4
例 4

Use the fact that the definite integrals of $f(x)$ and $g(x)$ have the following values with the lower and upper limits shown below:
已知函数 $f(x)$ 和 $g(x)$ 在所示上下限下的定积分值如下：

$$\begin{array}{rl}{\int }_0^{6}f(x)dx& =6\\ {\int }_0^{3}f(x)dx& =-2\\ {\int }_0^{6}g(x)dx& =-4\\ {\int }_0^{3}g(x)dx& =1\end{array}$$

Compute the following definite integrals.
计算下列定积分。

a. ${\int }_0^6[f(x)+g(x)]dx$

b. ${\int }_0^3[f(x)–g(x)]dx$

c. ${\int }_3^6[f(x)+g(x)]dx$

d. ${\int }_3^6[f(x)–g(x)]dx$

e. ${\int }_0^6[2f(x)+3g(x)]dx$

f. ${\int }_0^3[3f(x)–4g(x)]dx$

g. ${\int }_3^6[2f(x)+6g(x)]dx$

Solution
解

Apply the sum property of definite integrals,
应用定积分的和性质
${\int }_a^b[f(x)+g(x)]dx={\int }_a^{b}f(x)dx+{\int }_a^{b}g(x)dx$,

to determine the value of ${\int }_0^6[f(x)+g(x)]dx$.
来求 ${\int }_0^6[f(x)+g(x)]dx$ 的值。

$$\begin{array}{rl}{\int }_0^6[f(x)+g(x)]dx& ={\int }_0^{6}f(x)dx+{\int }_0^{6}g(x)dx\end{array}$$

Use the given values for the two definite integrals to find the value of ${\int }_0^6[f(x)+g(x)]dx$.
利用两个定积分的给定值来求 ${\int }_0^6[f(x)+g(x)]dx$ 的值。

a. This means that ${\int }_0^6[f(x)+g(x)]dx$ is equal to $2$.
a. 这意味着 ${\int }_0^6[f(x)+g(x)]dx=2$。

We’ll apply a similar process to evaluate the definite integral,
我们将用类似的方法来计算定积分
${\int }_0^3[f(x)–g(x)]dx$.

This time, we’ll use the difference property of definite integrals,
这次，我们将使用定积分的差性质：

${\int }_a^b[f(x)–g(x)]dx={\int }_a^{b}f(x)dx-{\int }_a^{b}g(x)dx$.

$$\begin{array}{rl}{\int }_0^6[f(x)–g(x)]dx& ={\int }_0^{6}f(x)dx–{\int }_0^{6}g(x)dx\\ & =6–(-4)\\ & =10\end{array}$$

b. Hence, ${\int }_0^6[f(x)–g(x)]dx$ is equal to $10$.
b. 因此，${\int }_0^6[f(x)–g(x)]dx=10$。

The given lower and upper limits of the definite integral, ${\int }_3^6[f(x)+g(x)]dx$, are the result of the four given definite integrals being combined. Rewrite ${\int }_3^6[f(x)+g(x)]dx$ in terms of the four given definite integrals and by applying the different properties for definite integrals.
定积分 ${\int }_3^6[f(x)+g(x)]dx$ 的给定上下限是由四个给定的定积分合并得到的。通过应用定积分的不同性质，用四个给定的定积分来重写 ${\int }_3^6[f(x)+g(x)]dx$。

• Use the sum property to distribute the definite integral operation.
  利用和性质分配定积分运算。

• Expand ${\int }_3^{6}f(x)dx$ and ${\int }_3^{6}g(x)dx$ by reversing the process of combining a definite integral’s intervals, ${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$.
  通过反转定积分区间的合并过程 ${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$，展开 ${\int }_3^{6}f(x)dx$ 和 ${\int }_3^{6}g(x)dx$。

• Reverse the limits of ${\int }_3^{0}f(x)dx$ and ${\int }_3^{0}g(x)dx$ using the reverse interval property, ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$.
  利用区间反转性质 ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$，交换 ${\int }_3^{0}f(x)dx$ 和 ${\int }_3^{0}g(x)dx$ 的上下限。

$$\begin{array}{rl}{\int }_3^6[f(x)+g(x)]dx& ={\int }_3^{6}f(x)dx+{\int }_3^{6}g(x)dx\\ & =\left[{\int }_3^{0}f(x)dx+{\int }_0^{6}f(x)dx\right]+\left[{\int }_3^{0}g(x)dx+{\int }_0^{6}g(x)dx\right]\\ & =\left[-{\int }_0^{3}f(x)dx+{\int }_0^{6}f(x)dx\right]+\left[-{\int }_0^{3}g(x)dx+{\int }_0^{6}g(x)dx\right]\\ & =\left[{\int }_0^{6}f(x)dx–{\int }_0^{3}f(x)dx\right]+\left[{\int }_0^{6}g(x)dx-{\int }_0^{3}g(x)dx\right]\end{array}$$

We’ve now expressed the definite integral in term of the following:
现在我们已经用以下定积分表示出了所求定积分：

$$\begin{array}{rl}{\int }_0^{6}f(x)dx& =6\\ {\int }_0^{3}f(x)dx& =-2\\ {\int }_0^{6}g(x)dx& =-4\\ {\int }_0^{3}g(x)dx& =1\end{array}$$

We’ll substitute these values into our last expression to find the value of ${\int }_3^6[f(x)+g(x)]dx$.
我们将这些值代入上面的表达式，求出 ${\int }_3^6[f(x)+g(x)]dx$ 的值。

$$\begin{array}{rl}{\int }_3^6[f(x)+g(x)]dx& =\left[{\int }_0^{6}f(x)dx–{\int }_0^{3}f(x)dx\right]+\left[{\int }_0^{6}g(x)dx-{\int }_0^{3}g(x)dx\right]\\ & =[6–(-2)]+[-4–(1)]\\ & =8–5\\ & =3\end{array}$$

c. This means that ${\int }_3^6[f(x)+g(x)]dx$ is equal to $3$.
c. 这意味着 ${\int }_3^6[f(x)+g(x)]dx=3$。

We’ll apply the same process to evaluate the fourth definite integral, ${\int }_3^6[f(x)–g(x)]dx$.
我们将用同样的方法来计算第四个定积分 ${\int }_3^6[f(x)–g(x)]dx$。

$$\begin{array}{rl}{\int }_3^6[f(x)–g(x)]dx& ={\int }_3^{6}f(x)dx–{\int }_3^{6}g(x)dx,\text{Difference Property}\\ & =\left[{\int }_3^{0}f(x)dx+{\int }_0^{6}f(x)dx\right]–\left[{\int }_3^{0}g(x)dx+{\int }_0^{6}g(x)dx\right],\text{Combining Interval}\\ & =\left[-{\int }_0^{3}f(x)dx+{\int }_0^{6}f(x)dx\right]–\left[-{\int }_0^{3}g(x)dx+{\int }_0^{6}g(x)dx\right],\text{Reverse Interval}\\ & =\left[{\int }_0^{6}f(x)dx–{\int }_0^{3}f(x)dx\right]–\left[{\int }_0^{6}g(x)dx-{\int }_0^{3}g(x)dx\right]\end{array}$$

Use the same values to determine the value of ${\int }_3^6[f(x)–g(x)]dx$.
用相同的值来确定 ${\int }_3^6[f(x)–g(x)]dx$ 的值。

$$\begin{array}{rl}{\int }_3^6[f(x)–g(x)]dx& =\left[{\int }_0^{6}f(x)dx–{\int }_0^{3}f(x)dx\right]–\left[{\int }_0^{6}g(x)dx-{\int }_0^{3}g(x)dx\right]\\ & =[6–(-2)]–[-4-(1)]\\ & =8–(-5)\\ & =13\end{array}$$

d. Hence, we have ${\int }_3^6[f(x)–g(x)]dx=13$.
d. 因此，${\int }_3^6[f(x)–g(x)]dx=13$。

Use the sum and difference property followed by the constant multiple property to evaluate the next two definite integrals. Once we’ve rewritten the two definite integrals, substitute the given values into our new expression.
先用和差性质，再用常数因子性质来计算接下来的两个定积分。重写这两个定积分后，将给定的值代入新的表达式中。

$$\begin{array}{r}{\int }_0^6[2\mathbf{f}(\mathbf{x})+3\mathbf{g}(\mathbf{x})]\mathbf{dx}\end{array}$$	$$\begin{array}{rl}{\int }_0^6[2f(x)+3g(x)]dx& ={\int }_0^{6}2f(x)dx+{\int }_0^{6}3g(x)dx,\text{Sum Property 和性质}\\ & =2{\int }_0^{6}f(x)dx+3{\int }_0^{6}g(x)dx,\text{Constant Multiple Property 常数因子性质}\\ & =2(6)+3(-4)\\ & =12–12\\ & =0\end{array}$$
$$\begin{array}{r}{\int }_0^3[3\mathbf{f}(\mathbf{x})–4\mathbf{g}(\mathbf{x})]\mathbf{dx}\end{array}$$	$$\begin{array}{rl}{\int }_0^3[3f(x)–4g(x)]dx& ={\int }_0^{3}3f(x)dx–{\int }_0^{3}4g(x)dx,\text{Difference Property 差性质}\\ & =3{\int }_0^{3}f(x)dx-4{\int }_0^{3}g(x)dx,\text{Constant Multiple Property 常数因子性质}\\ & =3(-2)–4(1)\\ & =-6–4\\ & =-10\end{array}$$

This shows that through different definite integral properties, we can manipulate definite integrals to evaluate them and find their actual values. For our case, we have the following values:
这表明，通过不同的定积分性质，我们可以对定积分进行变形，从而计算出它们的实际值。在我们的例子中，得到以下结果：

e. ${\int }_0^6[2f(x)+3g(x)]dx=0$

f. ${\int }_0^3[3f(x)–4g(x)]dx=-10$

The last definite integral that we need to evaluate is ${\int }_3^6[2f(x)+6g(x)]dx$. This is also a quick review to make sure you understand how we were able to evaluate and simplify the last five items.
我们需要计算的最后一个定积分是 ${\int }_3^6[2f(x)+6g(x)]dx$。这也是一个快速复习，以确保你理解我们是如何计算和简化前五个积分的。

• Apply the sum and constant multiple properties.
  应用和性质和常数因子性质。

• Rewrite the integrals so they lower and upper limits that we can work with.
  重写积分，使它们的上下限是我们可以处理的。

• Substitute the values that we know to simplify the expression and return an actual value.
  代入已知的值来简化表达式并得到实际结果。

$$\begin{array}{rl}{\int }_3^6[2f(x)+6g(x)]dx& ={\int }_3^{6}2f(x)dx+{\int }_3^{6}6g(x)dx,\text{Sum Property}\\ & =2{\int }_3^{6}f(x)dx+6{\int }_3^{6}g(x)dx,\text{Constant Multiple Property}\\ & =2\left[{\int }_3^{0}f(x)dx+{\int }_0^{6}f(x)dx\right]+6\left[{\int }_3^{0}g(x)dx+{\int }_0^{6}g(x)dx\right],\text{Combining Interval}\\ & =2\left[-{\int }_0^{3}f(x)dx+{\int }_0^{6}f(x)dx\right]+6\left[-{\int }_0^{3}g(x)dx+{\int }_0^{6}g(x)dx\right],\text{Reversing Interval}\\ & =2\left[{\int }_0^{6}f(x)dx-{\int }_0^{3}f(x)dx\right]+6\left[{\int }_0^{6}g(x)dx-{\int }_0^{3}g(x)dx\right]\\ & =2[6–(-2)]+6[-4–(1)]\\ & =2(8)+6(-5)\\ & =16–30\\ & =-14\end{array}$$

g. This means that ${\int }_3^6[2f(x)+6g(x)]dx$ is equal to $-14$.
g. 这意味着 ${\int }_3^6[2f(x)+6g(x)]dx=-14$。

Example 5
例 5

Show that ${\int }_0^1\sqrt{1+2x^4}dx<1.4$. Use the fact that $\int \sqrt{1+x}dx=\frac{1}{3}(1+2x{)}^{\frac{3}{2}}+C$.
证明 ${\int }_0^1\sqrt{1+2x^4}dx<1.4$。利用 $\int \sqrt{1+2x}dx=\frac{1}{3}(1+2x{)}^{\frac{3}{2}}+C$ 这一结果。

Solution
解

Since we’re working with definite integrals with lower and upper limits at $x=0$ and $x=1$. Let’s evaluate the definite integral, ${\int }_0^1\sqrt{1+2x}dx$ using the indefinite integral given from the problem.
由于我们要处理的是上下限为 $x=0$ 和 $x=1$ 的定积分。让我们利用题目中给出的不定积分来计算定积分 ${\int }_0^1\sqrt{1+2x}dx$。

$$\begin{array}{rl}{\int }_0^1\sqrt{1+2x}dx& =\frac{1}{3}(1+2x{)}^{\frac{3}{2}}{|}_0^1\\ & =\frac{1}{3}[(1+2{)}^{\frac{3}{2}}-(1+0{)}^{\frac{3}{2}}]\\ & =\frac{1}{3}(3^{\frac{3}{2}}–1)\\ & \approx 1.399\end{array}$$

When $x$ is within the interval, $[0,1]$, $x^4$ will be less than or equal to $x$. Let’s manipulate the inequality, $x^4\le x$, so that we end up comparing $\sqrt{1+2x^4}$ and $\sqrt{1+2x}$.
当 $x$ 在区间 $[0,1]$ 内时，$x^4$ 小于或等于 $x$。我们对不等式 $x^4\le x$ 进行变形，以便最终比较 $\sqrt{1+2x^4}$ 和 $\sqrt{1+2x}$。

$$\begin{array}{rl}{x}^4& \le x\\ 2x^4& \le 2x\\ 1+2x^4& \le 1+2x\\ \sqrt{1+2x^4}& \le \sqrt{1+2x}\end{array}$$

Now that we’ve shown that $\sqrt{1+2x^4}\le \sqrt{1+2x}$, we can use the comparison property to estimate its value. Recall that when $f(x)\le g(x)$ for $x\in [a,b]$, $\int_{a}^{b} f(x)dx \leq \int_{a}^{b} g(x)dx $.
既然我们已经证明了 $\sqrt{1+2x^4}\le \sqrt{1+2x}$，我们可以使用比较性质来估计它的值。回想一下，当对于 $x\in [a,b]$ 有 $f(x)\le g(x)$ 时，${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx$。

$$\begin{array}{rl}\sqrt{1+2x^4}& \le \sqrt{1+2x}\\ {\int }_0^1\sqrt{1+2x^4}dx& \le {\int }_0^1\sqrt{1+2x}dx\\ {\int }_0^1\sqrt{1+2x^4}dx& \le {\int }_0^1\sqrt{1+2x}dx<1.4\end{array}$$

Hence, we’ve just shown that
因此，我们刚刚证明了

${\int }_0^1\sqrt{1+2x^4}dx<1.4$.

Practice Questions

练习题

1. Use the fact that ${\int }_0^{8}f(x)dx=-5$ and ${\int }_{-8}^{0}f(x)dx=35$ to determine the value of ${\int }_{-8}^{8}f(x)dx$.
2. Use the fact that ${\int }_{-2}^{1}g(x)dx=10$ and ${\int }_1^{2}g(x)dx=-12$ to determine the value of ${\int }_{-2}^{2}g(x)dx$.
3. Use the fact that ${\int }_0^{5}f(x)dx=-4$ and ${\int }_0^{12}f(x)dx=-12$ to determine the value of ${\int }_5^{12}f(x)dx$.
4. Use the fact that ${\int }_{-2}^{4}g(x)dx=22$ and ${\int }_{-2}^{8}g(x)dx=-5$ to determine the value of ${\int }_4^{8}g(x)dx$.
5. Prove the zero-length interval property, ${\int }_a^{a}f(x)dx=0$.
6. Prove the difference property for indefinite integral, $\int [f(x)–g(x)]dx=\int f(x)dx–\int g(x)dx$.
7. Use the fact that the definite integrals of $f(x)$ and $g(x)$ have the following values with the lower and upper limits shown below:
   $$\begin{array}{rl}{\int }_0^{8}f(x)dx& =24\\ {\int }_0^{4}f(x)dx& =-12\\ {\int }_0^{8}g(x)dx& =-16\\ {\int }_0^{4}g(x)dx& =6\end{array}$$
   Compute the following definite integrals.
   a. ${\int }_0^8[f(x)+g(x)]dx$
   b. ${\int }_0^4[f(x)–g(x)]dx$
   c. ${\int }_4^8[f(x)+g(x)]dx$
   d. ${\int }_4^8[f(x)–g(x)]dx$
   e. ${\int }_0^8[4f(x)+g(x)]dx$
   f. ${\int }_0^4[6f(x)–5g(x)]dx$
   g. ${\int }_4^8[12f(x)–4g(x)]dx$
8. Use the fact that the definite integrals of $f(x)$ and $g(x)$ have the following values with the lower and upper limits shown below:
   $$\begin{array}{rl}{\int }_0^{10}f(x)dx& =-24\\ {\int }_0^{5}f(x)dx& =12\\ {\int }_0^{10}g(x)dx& =8\\ {\int }_0^{5}g(x)dx& =-6\end{array}$$
   Compute the following definite integrals.
   a. ${\int }_0^{10}[f(x)+g(x)]dx$
   b. ${\int }_0^5[f(x)–g(x)]dx$
   c. ${\int }_5^{10}[f(x)+g(x)]dx$
   d. ${\int }_5^{10}[f(x)–g(x)]dx$
   e. ${\int }_0^{10}[f(x)+6g(x)]dx$
   f. ${\int }_0^5[2f(x)–3g(x)]dx$
   g. ${\int }_5^{10}[3f(x)+g(x)]dx$
9. Show that ${\int }_0^1\sqrt{1+4x^3}dx<1.7$. Use the fact that $\int \sqrt{1+4x}dx=\frac{1}{6}(1+4x{)}^{\frac{3}{2}}+C$.
10. Show that ${\int }_0^1\sqrt{2+x^4}dx<1.6$. Use the fact that $\int \sqrt{2+x}dx=\frac{2}{3}(2+x{)}^{\frac{3}{2}}+C$.

Answer Key

答案

1. ${\int }_{-8}^{8}f(x)dx=30$

2. ${\int }_{-2}^{2}g(x)dx=-2$

3. ${\int }_5^{12}f(x)dx=-8$

4. ${\int }_4^{8}g(x)dx=-27$

5. If $F(x)$ is the antiderivative of $\int f(x)dx$, we have the following:
   $$\begin{array}{rl}{\int }_a^{b}f(x)dx& =F(b)-F(a)\\ & \\ \Rightarrow {\int }_a^{a}f(x)dx& =F(a)–F(a)\\ & =0\end{array}$$
   This confirms the zero-length interval property for definite integrals, ${\int }_a^{a}f(x)dx=0$.

6. $$\begin{array}{rl}\frac{d}{dx}[f(x)–g(x)]& =\frac{d}{dx}f(x)–\frac{d}{dx}g(x)\\ \frac{d}{dx}\int f(x)dx& =f(x)\\ & \\ \frac{d}{dx}\int [f(x)–g(x)]dx& =f(x)–g(x)\\ \frac{d}{dx}\left[\int f(x)dx-\int g(x)dx\right]& =\frac{d}{dx}\int f(x)dx-\frac{d}{dx}\int g(x)dx\\ & =f(x)–g(x)\\ \Rightarrow \int [f(x)–g(x)]dx& =\int f(x)dx–\int g(x)dx\end{array}$$

7. a. ${\int }_0^8[f(x)+g(x)]dx=8$

   b. ${\int }_0^4[f(x)–g(x)]dx=-18$

   c. ${\int }_4^8[f(x)+g(x)]dx=14$

   d. ${\int }_4^8[f(x)–g(x)]dx=58$

   e. ${\int }_0^8[4f(x)+g(x)]dx=80$

   f. ${\int }_0^4[6f(x)–5g(x)]dx=-102$

   g. ${\int }_4^8[12f(x)–4g(x)]dx=520$

8. a. ${\int }_0^{10}[f(x)+g(x)]dx=-16$

   b. ${\int }_0^5[f(x)–g(x)]dx=18$

   c. ${\int }_5^{10}[f(x)+g(x)]dx=-22$

   d. ${\int }_5^{10}[f(x)–g(x)]dx=-50$

   e. ${\int }_0^{10}[f(x)+6g(x)]dx=24$

   f. ${\int }_0^5[2f(x)–3g(x)]dx=42$

   g. ${\int }_5^{10}[3f(x)+g(x)]dx=-94$

9. $$\begin{array}{rl}{\int }_0^1\sqrt{1+4x}dx& \approx 1.697\\ & \\ x^3& \le x\\ \sqrt{1+4x^3}& \le \sqrt{1+4x}\\ {\int }_0^1\sqrt{1+4x^3}dx& \le {\int }_0^1\sqrt{1+4x}dx\\ {\int }_0^1\sqrt{1+4x^3}dx& <1.7\end{array}$$

10. $$\begin{array}{rl}{\int }_0^1\sqrt{2+x}dx& \approx 1.578\\ & \\ x^4& \le x\\ \sqrt{2+x^4}& \le \sqrt{2+x}\\ {\int }_0^1\sqrt{2+x^4}dx& \le {\int }_0^1\sqrt{2+x}dx\\ {\int }_0^1\sqrt{2+x^4}dx& <1.6\end{array}$$

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