$$ \int \int _ { R } \frac { 1 } { ( 2 x + 3 y ) ^ { 2 } } d A , R = [ 0,1 ] imes [ 1,2 ] $$

Question

$$ \int \int _ { R } \frac { 1 } { ( 2 x + 3 y ) ^ { 2 } } d A , R = [ 0,1 ] 	imes [ 1,2 ] $$

UpStudy AI · Accepted Answer

To solve the given double integral, we first need to find the limits of integration for the region $$ R $$ which is defined as $$ [0,1] \times [1,2] $$.

The double integral is given by:
$$ \int \int _ { R } \frac { 1 } { (2x + 3y)^2} dA $$

To solve this integral, we can use the following steps:

1. Find the limits of integration for the region $$ R $$.
2. Evaluate the integral with respect to $$ x $$ first, and then with respect to $$ y $$.

Let's start by finding the limits of integration for the region $$ R $$.
Evaluate by following steps:
- step0: Solution:
  $$\left\{ \begin{array}{l}0\leq x\x\leq 1\1\leq y\y\leq 2\end{array}\right.$$
- step1: Solve the inequality:
  $$\left\{ \begin{array}{l}x\geq 0\x\leq 1\y\geq 1\y\leq 2\end{array}\right.$$
- step2: Find the intersection:
  $$\left\{ \begin{array}{l}0\leq x\leq 1\1\leq y\leq 2\end{array}\right.$$
The limits of integration for the region $$ R $$ are:
$$ 0 \leq x \leq 1 $$
$$ 1 \leq y \leq 2 $$

Now, let's evaluate the integral with respect to $$ x $$ first, and then with respect to $$ y $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{1} \int_{1}^{2} \frac{1}{\left(2x+3y\right)^{2}} dy dx$$
- step1: Evaluate the inner integral:
  $$\int_{0}^{1} \left(-\frac{1}{6x+18}+\frac{1}{6x+9}\right) dx$$
- step2: Evaluate the integral:
  $$\int \left(-\frac{1}{6x+18}+\frac{1}{6x+9}\right) dx$$
- step3: Use properties of integrals:
  $$\int -\frac{1}{6x+18} dx+\int \frac{1}{6x+9} dx$$
- step4: Evaluate the integral:
  $$-\frac{1}{6}\ln{\left(\left|x+3\right|\right)}+\int \frac{1}{6x+9} dx$$
- step5: Evaluate the integral:
  $$-\frac{1}{6}\ln{\left(\left|x+3\right|\right)}+\frac{\ln{\left(\left|2x+3\right|\right)}}{6}$$
- step6: Return the limits:
  $$\left(-\frac{1}{6}\ln{\left(\left|x+3\right|\right)}+\frac{\ln{\left(\left|2x+3\right|\right)}}{6}\right)\bigg |_{0}^{1}$$
- step7: Calculate the value:
  $$\frac{\ln{\left(\frac{5}{4}\right)}}{6}$$
The double integral $$ \int \int _ { R } \frac { 1 } { (2x + 3y)^2} dA $$ over the region $$ R = [0,1] \times [1,2] $$ is equal to $$ \frac{\ln(5/4)}{6} $$ or $$ \frac{\ln(1\frac{1}{4})}{6} $$.
The value of the double integral is $$ \frac{\ln(5/4)}{6} $$ or $$ \frac{\ln(1\frac{1}{4})}{6} $$.

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