Math / Calculus

Questionate $\iint_{D} 2 d A$ , where $D=((x, y): 0 \leq x \leq 1, x \leq y \leq 1+x)$ .

Studdy Solution

STEP 1

1. The region $D$ is defined in terms of $x$ and $y$ with specific bounds.
2. The function to integrate is a constant function, $f(x, y) = 2$ .
3. We will use double integration over the specified region $D$ .

STEP 2

1. Understand the region of integration $D$ .
2. Set up the double integral with the correct limits of integration.
3. Evaluate the double integral.

STEP 3

Understand the region $D$ . The region $D$ is defined as:
$D = \{(x, y) \mid 0 \leq x \leq 1, \, x \leq y \leq 1+x\}$
This describes a region in the $xy$ -plane where: - $x$ ranges from 0 to 1. - For a fixed $x$ , $y$ ranges from $x$ to $1+x$ .

STEP 4

Set up the double integral with the correct limits of integration. The order of integration will be $dy \, dx$ :
$\iint_{D} 2 \, dA = \int_{0}^{1} \int_{x}^{1+x} 2 \, dy \, dx$

STEP 5

Evaluate the inner integral with respect to $y$ :
$\int_{x}^{1+x} 2 \, dy = 2[y]_{x}^{1+x} = 2((1+x) - x) = 2$

STEP 6

Evaluate the outer integral with respect to $x$ :
$\int_{0}^{1} 2 \, dx = 2[x]_{0}^{1} = 2(1 - 0) = 2$
The value of the double integral is:
$\boxed{2}$

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Questionate $\iint_{D} 2 d A$ , where $D=((x, y): 0 \leq x \leq 1, x \leq y \leq 1+x)$ .

Studdy Solution

STEP 1

1. The region $D$ is defined in terms of $x$ and $y$ with specific bounds.
2. The function to integrate is a constant function, $f(x, y) = 2$ .
3. We will use double integration over the specified region $D$ .

STEP 2

1. Understand the region of integration $D$ .
2. Set up the double integral with the correct limits of integration.
3. Evaluate the double integral.

STEP 3

Understand the region $D$ . The region $D$ is defined as:
$D = \{(x, y) \mid 0 \leq x \leq 1, \, x \leq y \leq 1+x\}$
This describes a region in the $xy$ -plane where: - $x$ ranges from 0 to 1. - For a fixed $x$ , $y$ ranges from $x$ to $1+x$ .

STEP 4

Set up the double integral with the correct limits of integration. The order of integration will be $dy \, dx$ :
$\iint_{D} 2 \, dA = \int_{0}^{1} \int_{x}^{1+x} 2 \, dy \, dx$

STEP 5

Evaluate the inner integral with respect to $y$ :
$\int_{x}^{1+x} 2 \, dy = 2[y]_{x}^{1+x} = 2((1+x) - x) = 2$

STEP 6

Evaluate the outer integral with respect to $x$ :
$\int_{0}^{1} 2 \, dx = 2[x]_{0}^{1} = 2(1 - 0) = 2$
The value of the double integral is:
$\boxed{2}$

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Questionate ∬D2dA\iint_{D} 2 d A∬D​2dA, where D=((x,y):0≤x≤1,x≤y≤1+x)D=((x, y): 0 \leq x \leq 1, x \leq y \leq 1+x)D=((x,y):0≤x≤1,x≤y≤1+x).

Studdy Solution

STEP 1

1. The region D D D is defined in terms of x x x and y y y with specific bounds.2. The function to integrate is a constant function, f(x,y)=2 f(x, y) = 2 f(x,y)=2.3. We will use double integration over the specified region D D D.

STEP 2

1. Understand the region of integration D D D.2. Set up the double integral with the correct limits of integration.3. Evaluate the double integral.

STEP 3

STEP 4

Set up the double integral with the correct limits of integration. The order of integration will be dy dx dy \, dx dydx:∬D2 dA=∫01∫x1+x2 dy dx\iint_{D} 2 \, dA = \int_{0}^{1} \int_{x}^{1+x} 2 \, dy \, dx∬D​2dA=∫01​∫x1+x​2dydx

STEP 5

Evaluate the inner integral with respect to y y y:∫x1+x2 dy=2[y]x1+x=2((1+x)−x)=2\int_{x}^{1+x} 2 \, dy = 2[y]_{x}^{1+x} = 2((1+x) - x) = 2∫x1+x​2dy=2[y]x1+x​=2((1+x)−x)=2

STEP 6

Evaluate the outer integral with respect to x x x:∫012 dx=2[x]01=2(1−0)=2\int_{0}^{1} 2 \, dx = 2[x]_{0}^{1} = 2(1 - 0) = 2∫01​2dx=2[x]01​=2(1−0)=2 The value of the double integral is:2 \boxed{2} 2​

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Questionate $\iint_{D} 2 d A$ , where $D=((x, y): 0 \leq x \leq 1, x \leq y \leq 1+x)$ .

1. The region $D$ is defined in terms of $x$ and $y$ with specific bounds.
2. The function to integrate is a constant function, $f(x, y) = 2$ .
3. We will use double integration over the specified region $D$ .

1. Understand the region of integration $D$ .
2. Set up the double integral with the correct limits of integration.
3. Evaluate the double integral.

Set up the double integral with the correct limits of integration. The order of integration will be $dy \, dx$ :
$\iint_{D} 2 \, dA = \int_{0}^{1} \int_{x}^{1+x} 2 \, dy \, dx$

Evaluate the inner integral with respect to $y$ :
$\int_{x}^{1+x} 2 \, dy = 2[y]_{x}^{1+x} = 2((1+x) - x) = 2$

Evaluate the outer integral with respect to $x$ :
$\int_{0}^{1} 2 \, dx = 2[x]_{0}^{1} = 2(1 - 0) = 2$
The value of the double integral is:
$\boxed{2}$