Question

may i get a little help with this? not really sure what to put where.......thanks

Answer 1

sure what do u need help with

Answer 2

sorry, took me a second to find where i'd stashed the screenshot. lol.

Answer 3

i know there's an x=-1, a y=-1 border and a i estimate y=(-5/4)x+2.75....could be 2.5....

Answer 4

sorry to keep bugging you, but you're always so helpful, @ganeshie8. can you help me again pretty please?

Answer 5

in part a) we have `dydx`
so you fix x value first, and vary y

Answer 6

shoot a vertical arrow cutting across the Region

Answer 7

y=-1 and y=3. it's in between those

Answer 8

the vertical arrow enters the Region at y = -1,
and leaves teh region at y = -5/4(x-3)-1

Answer 9

Notice that the entering a boundary is a constant flat line y = -1
however leaving boundary is a slanted line : y = -5/4(x-3)-1

Answer 10

how did you get that y=-5/4...line? i thought the line on the triangle was y=(-5/4)+2.75

Answer 11

m=slope =-5/4 and b=2.5 or 2.75 estimating

Answer 12

take two points on the line : (-1, 4) and (3, -1)

Answer 13

2.75 is right, you have just simplified..

Answer 14

\[\large \iint\limits_{R} f(x,y) ~dA = \int\limits_{-1}^3~\int\limits_{-1}^{-5/4x+2.75} f(x,y)~dy~dx\]

Answer 15

does that look good ?

Answer 16

ok. yeah. that makes sense. would the x be the same except -1<x<4?

Answer 17

x starts at -1 and ends at 3

Answer 18

oh yeah. that's what i meant. oops

Answer 19

think of slicing the region vertically

Answer 20

Nice explaining ^_^ Medal

Answer 21

-1<y<4