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

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## 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 which is related to the chain rule of differentiation and uses the relation
$$\int w(x) d x=\int\left[w(u) \frac{d u}{d x}\right] d x .$$
Another valuable technique is integration by parts, as described by the relation proved in frame 324 .
$$\int u d v=u v-\int v d u$$
Frequently a number of different integration procedures are used in a single problem as illustrated in frames 329-331.

The method of partial fractions (frame 332) involves splitting a function into a sum of fractions with simpler denominators that can be integrated by other methods.

## 数学代写|微积分代写Calculus代考|Area under a Curve and Definite Integrals

The area $A$ under the curve of a function $f(x)$ between $x=a$ and $x=b$ can be found by dividing the area into $N$ narrow strips parallel to the $\gamma$-axis, each of area $f\left(x_i\right) \Delta x$, and summing the strips. In the limit as the width of each strip approaches zero, the limit of the sum approaches the area under the curve.
$$A=\lim {\Delta x \rightarrow 0} \sum{i=1}^N f\left(x_i\right) \Delta x .$$
This infinite sum is called a definite integral and is denoted by
$$\int_a^b f(x) d x=\lim {\Delta x \rightarrow 0} \sum{i=1}^N f\left(x_i\right) \Delta x .$$
The area function $A(x)$ has the variable $x$ for the upper limit of the definite integral in frame 341 , and the integration variable is renamed $u$.
$$A(x)=\int_a^x f(u) d u .$$
The area function $A(x)$ is an antiderivative of the function $f(x)$ (frame 346), $A^{\prime}(x)=f(x)$. Therefore, it is an indefinite integral.
$$A(x)=\int_a^x f(u) d u=F(x)+c,$$
where $F(x)$ is any particular antiderivative of $f(x)$, and $c$ is an arbitrary constant. If we want to know the area bounded by $x=a$ and some value $x$, the constant $c$ can be evaluated by noting that if $x=a$, then the area is zero, so $A(a)=F(a)+c=0$ and $c=-F(a)$. Therefore, $A(x)=F(x)-F(a)$. The area under the curve between $x=a$ and $x=b$ is then
$$A(b)=A=\int_a^b f(x) d x=F(b)-F(a)$$

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

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

$$\int u d v=u v-\int v d u$$

## 数学代写|微积分代写Calculus代考|Area under a Curve and Definite Integrals

$$A=\lim \Delta x \rightarrow 0 \sum i=1^N f\left(x_i\right) \Delta x$$

$$\int_a^b f(x) d x=\lim \Delta x \rightarrow 0 \sum i=1^N f\left(x_i\right) \Delta x$$

$$A(x)=\int_a^x f(u) d u$$

$$A(x)=\int_a^x f(u) d u=F(x)+c$$

$$A(b)=A=\int_a^b f(x) d x=F(b)-F(a)$$

## MATLAB代写

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