Double integral:
$$\huge \int\limits_{}^{}\int\limits_{R}^{} 2x-4y^3~dA, R = [-5,4] \times [0,3]$$

Question

Double integral:
$$\huge \int\limits_{}^{}\int\limits_{R}^{} 2x-4y^3~dA, R = [-5,4] \times [0,3]$$

terenzreignz · Answer

Well... this doesn't look too bad...
x goes from -5 to 4, 
y goes from 0 to 3
am I right?

terenzreignz · Answer

Oi @shamil98 work with me here XD
$$\Large \iint\limits_R2x-4y^3 dA$$

shamil98 · Answer

Yeah. what would be the first step tho?

terenzreignz · Answer

Here... now that we have the limits down, the rest is easy (sort of)
$$\Large \int\limits_{y=0}^3 \ \int\limits_{x=-5}^42x-4y^3 dxdy$$

terenzreignz · Answer

Catch me so far?

shamil98 · Answer

Yeah.

terenzreignz · Answer

Now, much like with partial differentiation, you integrate this part first
$$\Large \int\limits_{y=0}^3 \ \color{blue}{\int\limits_{x=-5}^42x-4y^3 dx}dy$$
While for the time being, treating y as a constant.

shamil98 · Answer

A bit lost, what would you integrate -4y^3 as ? ...

terenzreignz · Answer

That entire thingy is just constant (with respect to x) so it's simply $\large -4xy^3$

shamil98 · Answer

Oh

terenzreignz · Answer

that was your problem, I assume? :D

luigi0210 · Answer

*following along*

shamil98 · Answer

yeah, i was wondering that lol

terenzreignz · Answer

So... got it from here?

shamil98 · Answer

Next , you integrate with respect to y ?.. or do you evaluate it as a definite integral ? just started this topic

terenzreignz · Answer

There are limits... so... definite integral? LOL

terenzreignz · Answer

$$\Large \int\limits_{y=0}^3 \ \color{blue}{\int\limits_{x=-5}^42x-4y^3 dx}dy$$

primeralph · Answer

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