# 数学代写|微积分代写Calculus代写|MTH2010

## 数学代写|微积分代写Calculus代写|Antiderivative, Integration, and the Indefinite Integral

The goal of this section is to learn some techniques for integration, sometimes called antidifferentiation.

In this section we generally designate a function by $f(x)$. The concept of the antiderivative is fundamental to the process of integration and is easily explained. When a function $F(x)$ is differentiated to give $f(x)=d F / d x$, then $F(x)$ is an antiderivative of $f(x)$, that is,
$$F^{\prime}(x)=f(x) .$$
This notation describes the defining property of the antiderivative, although only in terms of the derivative $F^{\prime}(x)$, not $F(x)$ itself.
The antiderivative is usually written in the form:
$$F(x)=\int f(x) d x .$$
The expression $\int f(x) d x$ is also called the integral of $f(x)$. The symbol $\int$ is known as the integration symbol; it represents the inverse of differentiation.
To summarize the notation, if $F^{\prime}(x)=f(x)$, then $F(x)$ is the or
$3 0 7 \longdiv { 4 }$ Go to 307.
Indefinite Integral:
Often one can find the antiderivative simply by guesswork. For instance, if $f(x)=1$, then $F(x)=\int f(x) d x=x$. To prove this, note that
$$F^{\prime}(x)=\frac{d}{d x}(x)=1=f(x) .$$
However, $f(x)$ is not the only antiderivative of $f(x)=1 ; x+c$, where $c$ is a constant, is also an antiderivative because
$$\frac{d}{d x}(x+c)=1+0=f(x) .$$

## 数学代写|微积分代写Calculus代写|Some Techniques of Integration

Often an unfamiliar function can be converted into a familiar function having a known integral by using a technique called change of variable. The method applies to integrating a “function of a function.” (Differentiation of such a function was discussed in frame 198. It is differentiated using the chain rule.) For example, $e^{-x^2}$ can be written $e^{-u}$, where $u=x^2$. With the following rule, the integral with respect to the variable $x$ can be converted into another integral, often simpler, depending on the variable $u$.
$$\int w(x) d x=\int\left[w(u) \frac{d x}{d u}\right] d u .$$
Let’s see how this works by applying it to a few problems.

Consider the problem of evaluating the integral
$$\int x e^{-x^2} d x$$
Let $u=x^2$, or $x=\sqrt{u}$, and $w(u)=\sqrt{u} e^{-u}$. Hence $\frac{d x}{d u}=\frac{1}{2 \sqrt{u}}$. Using the rule for change of variable, $\int w(x) d x=\int\left[u(u) \frac{d x}{d u}\right] d u$, the integral becomes
$$\int x e^{-u} \frac{1}{2 x} d u=\frac{1}{2} \int e^{-u} d u=-\frac{1}{2} e^{-u}+c=-\frac{1}{2} e^{-x^2}+c .$$
To prove that this result is correct, note that
$$\frac{d}{d x}\left(-\frac{1}{2} e^{-x^2}+c\right)=x e^{-x^2},$$
as required.
Try the following somewhat tricky problem. If you need a hint, see frame $\mathbf{3 1 7}$. Evaluate
$$\int \sin \theta \cos \theta d \theta=$$
To check your answer, go to $\mathbf{3 1 7}$.

# 微积分代考

## 数学代写|微积分代写微积分代写|不定积分，积分和不定积分

$$F^{\prime}(x)=f(x) .$$

$$F(x)=\int f(x) d x .$$

$3 0 7 \longdiv { 4 }$转到307。不定积分:通常人们可以简单地通过猜测求出不定积分。例如，如果是$f(x)=1$，那么是$F(x)=\int f(x) d x=x$。为了证明这一点，注意
$$F^{\prime}(x)=\frac{d}{d x}(x)=1=f(x) .$$

$$\frac{d}{d x}(x+c)=1+0=f(x) .$$

## 数学代写|微积分代写Calculus代写|Some Techniques of Integration

$$\int w(x) d x=\int\left[w(u) \frac{d x}{d u}\right] d u .$$

$$\int x e^{-x^2} d x$$

$$\int x e^{-u} \frac{1}{2 x} d u=\frac{1}{2} \int e^{-u} d u=-\frac{1}{2} e^{-u}+c=-\frac{1}{2} e^{-x^2}+c .$$

$$\frac{d}{d x}\left(-\frac{1}{2} e^{-x^2}+c\right)=x e^{-x^2},$$

$$\int \sin \theta \cos \theta d \theta=$$

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