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

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

Expand the integrand completely:

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

Then

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

4 0

2 years ago

Vicky downloaded to apps on her iPhone the first app was $599 and the second Apple was 14. $33 how much did Vicky spend on the a

Gala2k [10]

Answer:

573.33

Step-by-step explanation:

599+14.33

= 573.33

(if rounding then its 600+14.33) = 614.33

5 0

2 years ago

BRAINLIST PLEASEEEE HELP

Aleksandr [31]

Answer:

About 527.79

Step-by-step explanation:

formula is A=2πrh+2πr^2

6 0

3 years ago

Read 2 more answers

What is equally between 1200 and 1600 ​

What is equally between 1200 and 1600