QuestionCalculate the double integral $J=\int_{0}(x + xy - 2x) \, dx \, dy$ over the region defined by $y=-x-x^2$ and $y=-x$ .

Studdy Solution

STEP 1

Assumptions1. The integral is a double integral over the region D defined by the inequalities $y=-x-x^{}$ and $y=-x$ . . The integrand is $f(x, y) = x + xy -x$ .

STEP 2

First, we need to find the limits of integration for y. The limits of y are given by the inequalities $-x-x^{2} \leq y \leq -x$ .

STEP 3

The limits of integration for x are from0 to1. This is because the lines $y=-x$ and $y=-x-x^{2}$ intersect at x=0 and x=1.

STEP 4

We will first integrate with respect to y. The integral becomes $J=\int_{0}^{1}\int_{-x-x^{2}}^{-x}(x+x y-2 x) d y d x$

STEP 5

Now, we will integrate the function $f(x, y) = x + xy -2x$ with respect to y. $\int (x+xy-2x) dy = xy + \frac{1}{2}xy^{2} -2xy$

STEP 6

Next, we will evaluate the integral at the limits of y, $-x-x^{2}$ and $-x$ .
$[x(-x) + \frac{1}{2}x(-x)^{2} -2x(-x)] - [x(-x-x^{2}) + \frac{1}{2}x(-x-x^{2})^{2} -2x(-x-x^{2})]$

STEP 7

implify the expression.
$[-x^{2} + x^{3} +2x^{2}] - [-x^{2} + x^{3} + x^{4} -2x^{2} +2x^{3} +2x^{4}]$

STEP 8

Combine like terms.
$x^{4} -2x^{3} +3x^{2}$

STEP 9

Now, we will integrate this result with respect to x from to.
$\int_{}^{} (x^{4} -2x^{3} +3x^{2}) dx$

STEP 10

Integrate the function $x^{4} -2x^{3} +3x^{2}$ with respect to x.
$\frac{}{5}x^{5} - \frac{}{2}x^{4} + x^{3}$

STEP 11

Evaluate the integral at the limits of x,0 and.
$[\frac{}{5}()^{5} - \frac{}{}()^{4} + ()^{3}] - [\frac{}{5}(0)^{5} - \frac{}{}(0)^{4} + (0)^{3}]$

STEP 12

implify the expression.
$\frac{}{5} - \frac{}{2} + = \frac{7}{10}$ So, $J=\frac{7}{10}$ .

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QuestionCalculate the double integral $J=\int_{0}(x + xy - 2x) \, dx \, dy$ over the region defined by $y=-x-x^2$ and $y=-x$ .

Studdy Solution

STEP 1

Assumptions1. The integral is a double integral over the region D defined by the inequalities $y=-x-x^{}$ and $y=-x$ . . The integrand is $f(x, y) = x + xy -x$ .

STEP 2

First, we need to find the limits of integration for y. The limits of y are given by the inequalities $-x-x^{2} \leq y \leq -x$ .

STEP 3

The limits of integration for x are from0 to1. This is because the lines $y=-x$ and $y=-x-x^{2}$ intersect at x=0 and x=1.

STEP 4

We will first integrate with respect to y. The integral becomes $J=\int_{0}^{1}\int_{-x-x^{2}}^{-x}(x+x y-2 x) d y d x$

STEP 5

Now, we will integrate the function $f(x, y) = x + xy -2x$ with respect to y. $\int (x+xy-2x) dy = xy + \frac{1}{2}xy^{2} -2xy$

STEP 6

Next, we will evaluate the integral at the limits of y, $-x-x^{2}$ and $-x$ .
$[x(-x) + \frac{1}{2}x(-x)^{2} -2x(-x)] - [x(-x-x^{2}) + \frac{1}{2}x(-x-x^{2})^{2} -2x(-x-x^{2})]$

STEP 7

implify the expression.
$[-x^{2} + x^{3} +2x^{2}] - [-x^{2} + x^{3} + x^{4} -2x^{2} +2x^{3} +2x^{4}]$

STEP 8

Combine like terms.
$x^{4} -2x^{3} +3x^{2}$

STEP 9

Now, we will integrate this result with respect to x from to.
$\int_{}^{} (x^{4} -2x^{3} +3x^{2}) dx$

STEP 10

Integrate the function $x^{4} -2x^{3} +3x^{2}$ with respect to x.
$\frac{}{5}x^{5} - \frac{}{2}x^{4} + x^{3}$

STEP 11

Evaluate the integral at the limits of x,0 and.
$[\frac{}{5}()^{5} - \frac{}{}()^{4} + ()^{3}] - [\frac{}{5}(0)^{5} - \frac{}{}(0)^{4} + (0)^{3}]$

STEP 12

implify the expression.
$\frac{}{5} - \frac{}{2} + = \frac{7}{10}$ So, $J=\frac{7}{10}$ .

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QuestionCalculate the double integral J=∫0(x+xy−2x) dx dyJ=\int_{0}(x + xy - 2x) \, dx \, dyJ=∫0​(x+xy−2x)dxdy over the region defined by y=−x−x2y=-x-x^2y=−x−x2 and y=−xy=-xy=−x.

Studdy Solution

STEP 1

Assumptions1. The integral is a double integral over the region D defined by the inequalities y=−x−xy=-x-x^{}y=−x−x and y=−xy=-xy=−x. . The integrand is f(x,y)=x+xy−xf(x, y) = x + xy -xf(x,y)=x+xy−x.

STEP 2

First, we need to find the limits of integration for y. The limits of y are given by the inequalities −x−x2≤y≤−x-x-x^{2} \leq y \leq -x−x−x2≤y≤−x.

STEP 3

The limits of integration for x are from0 to1. This is because the lines y=−xy=-xy=−x and y=−x−x2y=-x-x^{2}y=−x−x2 intersect at x=0 and x=1.

STEP 4

We will first integrate with respect to y. The integral becomesJ=∫01∫−x−x2−x(x+xy−2x)dydxJ=\int_{0}^{1}\int_{-x-x^{2}}^{-x}(x+x y-2 x) d y d xJ=∫01​∫−x−x2−x​(x+xy−2x)dydx

STEP 5

Now, we will integrate the function f(x,y)=x+xy−2xf(x, y) = x + xy -2xf(x,y)=x+xy−2x with respect to y.∫(x+xy−2x)dy=xy+12xy2−2xy\int (x+xy-2x) dy = xy + \frac{1}{2}xy^{2} -2xy∫(x+xy−2x)dy=xy+21​xy2−2xy

STEP 6

STEP 7

implify the expression.[−x2+x3+2x2]−[−x2+x3+x4−2x2+2x3+2x4][-x^{2} + x^{3} +2x^{2}] - [-x^{2} + x^{3} + x^{4} -2x^{2} +2x^{3} +2x^{4}][−x2+x3+2x2]−[−x2+x3+x4−2x2+2x3+2x4]

STEP 8

Combine like terms.x4−2x3+3x2x^{4} -2x^{3} +3x^{2}x4−2x3+3x2

STEP 9

Now, we will integrate this result with respect to x from to.∫(x4−2x3+3x2)dx\int_{}^{} (x^{4} -2x^{3} +3x^{2}) dx∫​(x4−2x3+3x2)dx

STEP 10

Integrate the function x4−2x3+3x2x^{4} -2x^{3} +3x^{2}x4−2x3+3x2 with respect to x.5x5−2x4+x3\frac{}{5}x^{5} - \frac{}{2}x^{4} + x^{3}5​x5−2​x4+x3

STEP 11

Evaluate the integral at the limits of x,0 and.[5()5−()4+()3]−[5(0)5−(0)4+(0)3][\frac{}{5}()^{5} - \frac{}{}()^{4} + ()^{3}] - [\frac{}{5}(0)^{5} - \frac{}{}(0)^{4} + (0)^{3}][5​()5−​()4+()3]−[5​(0)5−​(0)4+(0)3]

STEP 12

implify the expression.5−2+=710\frac{}{5} - \frac{}{2} + = \frac{7}{10}5​−2​+=107​So, J=710J=\frac{7}{10}J=107​.

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QuestionCalculate the double integral $J=\int_{0}(x + xy - 2x) \, dx \, dy$ over the region defined by $y=-x-x^2$ and $y=-x$ .

Assumptions1. The integral is a double integral over the region D defined by the inequalities $y=-x-x^{}$ and $y=-x$ . . The integrand is $f(x, y) = x + xy -x$ .

First, we need to find the limits of integration for y. The limits of y are given by the inequalities $-x-x^{2} \leq y \leq -x$ .

The limits of integration for x are from0 to1. This is because the lines $y=-x$ and $y=-x-x^{2}$ intersect at x=0 and x=1.

We will first integrate with respect to y. The integral becomes $J=\int_{0}^{1}\int_{-x-x^{2}}^{-x}(x+x y-2 x) d y d x$

Now, we will integrate the function $f(x, y) = x + xy -2x$ with respect to y. $\int (x+xy-2x) dy = xy + \frac{1}{2}xy^{2} -2xy$

implify the expression.
$[-x^{2} + x^{3} +2x^{2}] - [-x^{2} + x^{3} + x^{4} -2x^{2} +2x^{3} +2x^{4}]$

Combine like terms.
$x^{4} -2x^{3} +3x^{2}$

Now, we will integrate this result with respect to x from to.
$\int_{}^{} (x^{4} -2x^{3} +3x^{2}) dx$

Integrate the function $x^{4} -2x^{3} +3x^{2}$ with respect to x.
$\frac{}{5}x^{5} - \frac{}{2}x^{4} + x^{3}$

Evaluate the integral at the limits of x,0 and.
$[\frac{}{5}()^{5} - \frac{}{}()^{4} + ()^{3}] - [\frac{}{5}(0)^{5} - \frac{}{}(0)^{4} + (0)^{3}]$

implify the expression.
$\frac{}{5} - \frac{}{2} + = \frac{7}{10}$ So, $J=\frac{7}{10}$ .