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

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

In fact, a constant can always be added to a function without changing its derivative. If $y=F(x)$ is an antiderivative of $f(x)$, then all the antiderivatives of $f(x)$ can be denoted by writing
$$\int f(x) d x=F(x)+c,$$
where $c$ is an arbitrary constant.
If it is useful, we can write the defining equation for the antiderivative in terms of a differential: $d F=f(x) d x$. (If you need to review differentials, see frame 264.) Then we can describe all the antiderivatives by the notation
$$\int d F(x)=F(x)+c .$$
Because of the arbitrary constant, the definition is imprecise. For this reason this integral is called the indefinite integral.

In summary, the integral of the differential of a function is equal to the function plus a constant.

## 数学代写|微积分代写Calculus代考|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, $\int w(x) d x=\int\left[w(u) \frac{d x}{d u}\right] d u$, 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.

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

$$\int f(x) d x=F(x)+c$$

$$\int d F(x)=F(x)+c .$$

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

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

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

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

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

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