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# 数学代写|微积分代写Calculus代考|MAST10006 THE CONCEPT OF ANTIDERIVATIVE

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## 数学代写|微积分代写Calculus代考|THE CONCEPT OF ANTIDERIVATIVE

Let $f$ be a given function. We have already seen in the theory of falling bodies (Section 3.4) that it can be useful to find a function $F$ such that $F^{\prime}=f$. We call such a function $F$ an antiderivative of $f$. In fact we often want to find the most general function $F$, or a family of functions, whose derivative equals $f$. We can sometimes achieve this goal by a process of organized guessing.

Suppose that $f(x)=\cos x$. If we want to guess an antiderivative, then we are certainly not going to try a polynomial. For if we differentiate a polynomial then we get another polynomial. So that will not do the job. For similar reasons we are not going to guess a logarithm or an exponential. In fact the way that we get a trigonometric function through differentiation is by differentiating another trigonometric function. What trigonometric function, when differentiated, gives $\cos x$ ? There are only six functions to try, and a moment’s thought reveals that $F(x)=\sin x$ does the trick. In fact an even better answer is $F(x)=\sin x+C$. The constant differentiates to 0 , so $F^{\prime}(x)=f(x)=\cos x$. We have seen in our study of falling bodies that the additive constant gives us a certain amount of flexibility in solving problems.

Now suppose that $f(x)=x^2$. We have already noted that the way to get a polynomial through differentiation is to differentiate another polynomial. Since differentiation reduces the degree of the polynomial by 1 , it is natural to guess that the $F$ we seek is a polynomial of degree 3 . What about $F(x)=x^3$ ? We calculate that $F^{\prime}(x)=3 x^2$. That does not quite work. We seek $x^2$ for our derivative, but we got $3 x^2$. This result suggests adjusting our guess. We instead try $F(x)=x^3 / 3$. Then, indeed, $F^{\prime}(x)=3 x^2 / 3=x^2$, as desired. We will write $F(x)=x^3 / 3+C$ for our antiderivative.

More generally, suppose that $f(x)=a x^3+b x^2+c x+d$. Using the reasoning in the last paragraph, we may find fairly easily that $F(x)=a x^4 / 4+b x^3 / 3+$ $c x^2 / 2+d x+e$. Notice that, once again, we have thrown in an additive constant.

## 数学代写|微积分代写Calculus代考|THE INDEFINITE INTEGRAL

In practice, it is useful to have a compact notation for the antiderivative. What we do, instead of saying that “the antiderivative of $f(x)$ is $F(x)+C$,” is to write
$$\int f(x) d x=F(x)+C$$
So, for example,
$$\int \cos x d x=\sin x+C$$
and
$$\int x^3+x d x=\frac{x^4}{4}+\frac{x^2}{2}+C$$

and
$$\int e^{2 x} d x=\frac{e^{2 x}}{2}+C .$$
The symbol $\int$ is called an integral sign (the symbol is in fact an elongated “S”) and the symbol ” $d x$ ” plays a traditional role to remind us what the variable is. We call an expression like
$$\int f(x) d x$$
an indefinite integral. The name comes from the fact that later on we will have a notion of “definite integral” that specifies what value $C$ will take-so it is more definite in the answer that it gives.

## 数学代写|微积分他写Calculus他考|THE INDEFINITE INTEGRAL

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

$$\int \cos x d x=\sin x+C$$

$$\int x^3+x d x=\frac{x^4}{4}+\frac{x^2}{2}+C$$

$$\int e^{2 x} d x=\frac{e^{2 x}}{2}+C$$

$$\int f(x) d x$$

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