Help not much time.

Mathematics

2 answers:

Arisa [49]2 years ago

4 0

Answer:

(18.0)

Step-by-step explanation:

JulijaS [17]2 years ago

4 0

Answer:

Cross out no solution because there is no possible way for a math problem to have no solution if a problem were to have a infinite solution there wouldn’t be a solution if it was infinite so the answer i (0, 18)

Step-by-step explanation:

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ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$