## 数学代写|微积分代写Calculus代写|MTH125

2023年1月5日

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## 数学代写|微积分代写Calculus代写|Antiderivative rules

We learned to take derivatives of polynomials by learning the formula for the derivative of $x^n$ and then learning that (1) the derivative can be taken term-by-term and (2) constant multiples can be carried along to the derivative. The approach to finding antiderivatives of polynomials is much the same.

We start by seeking a power rule for antiderivatives. Perhaps reversing a derivative of a power helps. Reversing the derivative exercise
$$\frac{d}{d x} x^4=4 x^3$$
gives the antiderivative result
$$\int 4 x^3 d x=x^4+C .$$
We want a power rule for antidifferentiation, so it would be nice to know $\int x^3 d x$ instead. To get a function with a derivative that is $x^3$, dividing the original function by 4 does the trick:
$$\frac{d}{d x}\left(\frac{x^4}{4}\right)=\frac{d}{d x}\left(\frac{1}{4} \cdot x^4\right)=\frac{1}{4} \cdot 4 x^3=x^3 .$$
Reversing this exercise gives the antidifferentiation result
$$\int x^3 d x=\frac{x^4}{4}+C$$

Notice that going from $x^3$ to $\frac{x^4}{4}$ requires adding 1 to the exponent (changing from 3 to 4 ) and then dividing by the new exponent (dividing by 4). This process works in general. Because
$$\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=\frac{1}{n+1} \cdot(n+1) x^n=x^n,$$
the antiderivative power rule is
$$\int x^n d x=\frac{x^{n+1}}{n+1}+C$$

## 数学代写|微积分代写Calculus代写|Manipulating integrands

We have now reversed several derivative rules and given their antiderivative analogs, including the power rule, sum rule, difference rule, and constant multiple rule. Other derivative rules, such as the product rule, quotient rule, and chain rule, are more difficult to reverse. Although they have their analogs in antidifferentiation, they are not as straightforward to apply as recognizing a product or a quotient and using the associated rule. For this reason, we study them in chapter 7. In the meantime, if we wish to find antiderivatives of products, quotients, or compositions, we must change the form of the expression into something we can antidifferentiate using the rules we have at our disposal.
Example 10 Evaluate $\int x\left(x^2-3 x\right) d x$
Solution Because we have no product rule for antidifferentiation, we cannot evaluate the antiderivative in its current form. Instead, let’s perform the multiplication in the integrand:
$$\int x\left(x^2-3 x\right) d x=\int\left(x^3-3 x^2\right) d x$$

Now the integrand is in a form we can handle using the available rules; there is no product in the integral on the right. Using the antiderivative difference rule, the antiderivative constant multiple rule, and the antiderivative power rule, we have
\begin{aligned} \int\left(x^3-3 x^2\right) d x & =\frac{x^4}{4}-3 \cdot \frac{x^3}{3}+C \ & =\frac{x^4}{4}-x^3+C \end{aligned}
Example 11 Evaluate $\int \frac{x^3-4 x}{2 x^3} d x$.
Solution Because we have no quotient rule for integration, we cannot evaluate the integral in its current form. We need to manipulate the expression inside the integral (the integrand) into a form for which we can use the available rules. Because there is only one term in the denominator, we may perform the division easily:
$$\int \frac{x^5-4 x}{2 x^3} d x=\int\left(\frac{1}{2} x^2-2 x^{-2}\right) d x$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Antiderivative rules

$$\frac{d}{d x} x^4=4 x^3$$

$$\int 4 x^3 d x=x^4+C$$

$$\frac{d}{d x}\left(\frac{x^4}{4}\right)=\frac{d}{d x}\left(\frac{1}{4} \cdot x^4\right)=\frac{1}{4} \cdot 4 x^3=x^3$$

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

$$\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=\frac{1}{n+1} \cdot(n+1) x^n=x^n$$

$$\int x^n d x=\frac{x^{n+1}}{n+1}+C$$

## 数学代写|微积分代写Calculus代写|Manipulating integrands

$$\int x\left(x^2-3 x\right) d x=\int\left(x^3-3 x^2\right) d x$$

$$\int\left(x^3-3 x^2\right) d x=\frac{x^4}{4}-3 \cdot \frac{x^3}{3}+C=\frac{x^4}{4}-x^3+C$$

$$\int \frac{x^5-4 x}{2 x^3} d x=\int\left(\frac{1}{2} x^2-2 x^{-2}\right) d x$$

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## MATLAB代写

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