Double Int: int int [ 3v + 2u^2 + (6 - 3u - 2v)^2 ]

[shaoen01]

Hi,

I have been trying and working out this double integration for so many times, but i can't seem to get the answer 31. This step below is the last step that i have checked with the answer that i have managed to obtain correctly. From here onwards, it is downhill.

$\displaystyle \int_0^{2}\, \int_0^{\frac{6-3u}{2}}\, 3v\, +\, 2u^{2}\, + (6\, -\, 3u\, -\, 2v)^{2}$

 

[tkhunny]

Have you considered squaring the trinomial and doing all the parts separately?

 

[shaoen01]

Have you considered squaring the trinomial and doing all the parts separately?

Trinomial? I am not sure if i have learned that before, or i don't recall at least.

 

[stapel]

Trinomial? I am not sure if i have learned that before, or i don't recall at least.

Polynomials, and how to multiply them, should have been covered (at length) back in algebra.

There are many lessons online from which you can quickly learn the multiplication process. Polynomials are the sums of terms containing variables and numbers multiplied together.

Eliz.

 

[shaoen01]

Trinomial? I am not sure if i have learned that before, or i don't recall at least.

Polynomials, and how to multiply them, should have been covered (at length) back in algebra.

There are many lessons online from which you can quickly learn the multiplication process. Polynomials are the sums of terms containing variables and numbers multiplied together.

Eliz.

Hi Eliz,

Sorry i have always known trinomials as polynomials. I have been doing so much maths that everything is starting to get blur.

 

[stapel]

i have always known trinomials as polynomials.

"Tri-nomials" are just "poly-nomials" with three ("tri") terms.

...and I totally understand the "blur" thing....

Eliz.