How Do You Calculate the Area Between Three Curves?

[Chas3down]

Find region between these 3 curves.
2y=4sqrt(x)
y=5
2y+2x=6

Not sure how to find limits or actually setup the intigration.. is it just (left-mid-right?) But no idea for limits..

 

[SteamKing]

When in doubt, make a sketch of the three curves (actually, one curve and two lines) and see what drops out.

 

[Chas3down]

Hmm, I tried that and got a somewhat triangle, broke it up and got 2 integrals..
From -2 to 1
(5-(3-x))

From 1 to 6
5-2sqrt(x)

But it was wrong

 

[Mark44]

Find region between these 3 curves.
2y=4sqrt(x)
y=5
2y+2x=6

Not sure how to find limits or actually setup the intigration.. is it just (left-mid-right?) But no idea for limits..

You should simplify the 1st and 3rd equations.
y = 2√x
y + x = 3

There's no point in leaving in those common factors.

Hmm, I tried that and got a somewhat triangle, broke it up and got 2 integrals..
From -2 to 1
(5-(3-x))

From 1 to 6
5-2sqrt(x)

But it was wrong

Yes. You have the x coordinates for two of the intersection points correct, but the square root function and the horizontal line don't intersect for x = 6.

Try again.

 

[lurflurf]

almost
From -2 to 1
(5-(3-x))

From 1 to (try again)
5-2sqrt(x)
$$\mathrm{Area}=\int_{-2}^1 \! (5-(3-x))\, \mathrm{d}x+\int_1^{\text{try again}} \! (5-2\sqrt{x}) \, \mathrm{d}x$$