# 微积分作业代写calclulus代考| Change of variables in R2

my-assignmentexpert™ 微积分calculus作业代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。my-assignmentexpert™， 最高质量的微积分calculus作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于economics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此微积分calculus作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在经济学作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的微积分calculus代写服务。我们的专家在微积分calculus学 代写方面经验极为丰富，各种微积分calculus相关的作业也就用不着 说。

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

## 微积分作业代写calclulus代考|Change of variables in single integrals

Aim: We want to evaluate $\int_{a}^{b} f(x) \mathrm{d} x$ by invoking a transform $x=x(t)$.
Suppoosè $x(t)$ is strictly incréasing âs in Figuré $4.17(\mathrm{a})$, thèn $x^{\prime}(t)>0$, and $\frac{\mathrm{d} F}{\mathrm{~d} x}=f .$ Then
\begin{aligned} \int_{\alpha}^{\beta} f(x(t)) x^{\prime}(t) \mathrm{d} t &=\int_{\alpha}^{\beta} F^{\prime}(x(t)) x^{\prime}(t) \mathrm{d} t \ &=\int_{\alpha}^{\beta} \frac{\mathrm{d}}{\mathrm{d} t} F(x(t)) \mathrm{d} t=F(x(\beta))-F(x(\alpha)) \ &=F(b)-F(a)=\int_{a}^{b} f(x) \mathrm{d} x \end{aligned}
On the other hand suppose $x(t)$ is strictly decreasing as in the case shown in

Figure $4.17(\mathrm{~b})$, then $x^{\prime}(t)<0$, and $\frac{\mathrm{d} F}{\mathrm{~d} x}=f$. Then
\begin{aligned} \int_{\alpha}^{\beta} f(x(t))\left(-x^{\prime}(t)\right) \mathrm{d} t &=-\int_{\alpha}^{\beta} F^{\prime}(x(t)) x^{\prime}(t) \mathrm{d} t \ &=\int_{\beta}^{\alpha} \frac{\mathrm{d}}{\mathrm{d} t} F(x(t)) \mathrm{d} t=F(x(\alpha))-F(x(\beta)) \ &=F(b)-F(a)=\int_{a}^{b} f(x) \mathrm{d} x \end{aligned}
Consequently,
$$\int_{a}^{b} f(x) \mathrm{d} x=\int_{\alpha}^{\beta} f(x(t))\left|\frac{\mathrm{d} x}{\mathrm{~d} t}\right| \mathrm{d} t$$
So, in this case we see the integration interval has changed, but more significantly the change of variable $x \longrightarrow t$ has introduced a positive factor $\left|x^{\prime}\right|$ in the integral. This is called the scale factor since it scales up or down the interval size. We should expect a similar factor to appear in multiple integrals.

## 微积分作业代写calclulus代考|Change of variables in double integrals

For convenience we shall consider only bijective transformations:
$$\tau: \boldsymbol{u} \mapsto \boldsymbol{x}(\boldsymbol{u})=\left{\begin{array}{l} x=x(u, v) \ y=y(u, v) \end{array}\right.$$
such that $\frac{\partial(x, y)}{\partial(u, v)} \neq 0$. The Jacobian determinant (Definition $2.9$ ) is involved in the transformation of double integrals:
$\iint_{D} f(x, y) \mathrm{d} A \quad$ becomes expressed as $\iint_{E} g(u, v) \mathrm{d} A^{\prime}$
Geometrically the transformation affects areas both globally and locally.
To see how the transformation does this consider in Figure $4.18$ the “parallelogram” in the $x y$-plane created by constant $u$ and $v$ contours. Suppose opposite sides of the parallelogram are separated by differences $\mathrm{d} u$ and $\mathrm{d} v$.

For small $\mathrm{d} u$ and $\mathrm{d} v$, the area of the element in $D$ is given by the vector product
$\mathrm{d} A=|\overrightarrow{R P} \times \overrightarrow{R S}| \quad$ – geometric interpretation (see Page 3)
$=\left|\left(\mathrm{d} x_{u} e_{1}+\mathrm{d} y_{u} e_{2}\right) \times\left(\mathrm{d} x_{v} e_{1}+\mathrm{d} y_{v} e_{2}\right)\right| .$
$\underbrace{\text { along a line of }}{\text {along a line of }}$ constant $v \quad$ constant $u$ Thus, \begin{aligned} \mathrm{d} A &=\left|\left(\frac{\partial x}{\partial u} \mathrm{~d} u e{1}+\frac{\partial y}{\partial u} \mathrm{~d} u e_{2}\right) \times\left(\frac{\partial x}{\partial v} \mathrm{~d} v e_{1}+\frac{\partial y}{\partial v} \mathrm{~d} v e_{2}\right)\right| \ &=\left|\frac{\partial(x, y)}{\partial(u, v)}\right| \mathrm{d} u \mathrm{~d} v \quad \text { – by the chain rule (Section 2.G) product gives the Jacobian determinant } \end{aligned}
The reader should make particular note that it is the absolute value of the Jacobian determinant that appears here. This is reasonable since what we have done is transformed one area element to another, preserving the sign.

## 微积分作业代写calclulus代考|Change of variables in single integrals

∫一种bF(X(吨))X′(吨)d吨=∫一种bF′(X(吨))X′(吨)d吨 =∫一种bdd吨F(X(吨))d吨=F(X(b))−F(X(一种)) =F(b)−F(一种)=∫一种bF(X)dX

∫一种bF(X(吨))(−X′(吨))d吨=−∫一种bF′(X(吨))X′(吨)d吨 =∫b一种dd吨F(X(吨))d吨=F(X(一种))−F(X(b)) =F(b)−F(一种)=∫一种bF(X)dX

∫一种bF(X)dX=∫一种bF(X(吨))|dX d吨|d吨

## 微积分作业代写calclulus代考|Change of variables in double integrals

$$\tau: \boldsymbol{u} \mapsto \boldsymbol{x}(\boldsymbol{u})=\left{X=X(你,v) 和=和(你,v)\正确的。$$

∬DF(X,和)d一种变为表示为∬和G(你,v)d一种′

d一种=|R磷→×R小号→|– 几何解释（见第 3 页）
=|(dX你和1+d和你和2)×(dXv和1+d和v和2)|.
沿着一条线 ⏟沿着一条线 持续的v持续的你因此，d一种=|(∂X∂你 d你和1+∂和∂你 d你和2)×(∂X∂v dv和1+∂和∂v dv和2)| =|∂(X,和)∂(你,v)|d你 dv – 通过链式法则（第 2.G 节）乘积给出雅可比行列式