Question

∫∫y/x dxdy; D: 1≤x≤e, 4≤y≤6

Answer 1

Ответ:

9djeodkdkjjwejdddkekelrlell

Answer 2

To evaluate the double integral ∫∫(y/x) dxdy over the region D: 1≤x≤e, 4≤y≤6, we need to integrate the function (y/x) with respect to x and then with respect to y over the given limits.

Let's start by integrating with respect to x first:

∫(y/x) dx = y * ∫(1/x) dx

Now, integrate (1/x) with respect to x:

∫(1/x) dx = ln|x| + C

where C is the constant of integration.

Now, substitute the result back into the original double integral:

∫∫(y/x) dxdy = ∫(y * (ln|x| + C)) dy

Now, integrate with respect to y:

∫(y * (ln|x| + C)) dy = ∫(y * ln|x|) dy + ∫(y * C) dy

Integrate both parts:

∫(y * ln|x|) dy = (1/2)y^2 * ln|x| + D1 ∫(y * C) dy = C/2 * y^2 + D2

where D1 and D2 are constants of integration.

Now, combine the results:

∫∫(y/x) dxdy = (1/2)y^2 * ln|x| + C/2 * y^2 + D1 + D2

Now, evaluate the definite integral over the given limits:

∫∫(y/x) dxdy = [(1/2)y^2 * ln|x| + C/2 * y^2] evaluated from y=4 to y=6

Let's find the value of the definite integral:

∫∫(y/x) dxdy = [(1/2)(6)^2 * ln|e| + C/2 * (6)^2] - [(1/2)(4)^2 * ln|e| + C/2 * (4)^2]

∫∫(y/x) dxdy = [(1/2)(36) * ln(e) + C/2 * (36)] - [(1/2)(16) * ln(e) + C/2 * (16)]

Since ln(e) is equal to 1, the expression simplifies to:

∫∫(y/x) dxdy = [(1/2)(36) + C/2 * (36)] - [(1/2)(16) + C/2 * (16)]

∫∫(y/x) dxdy = [18 + 18C] - [8 + 8C]

Now, the constant terms cancel out:

∫∫(y/x) dxdy = 18 + 18C - 8 - 8C

Finally, the result is:

∫∫(y/x) dxdy = 10 + 10C

Therefore, the value of the given double integral over the region D: 1≤x≤e, 4≤y≤6 is 10 + 10C, where C is the constant of integration. Note that without knowing the specific value of C, we cannot obtain a numerical result.