Consider the area between the graphs x+2y=9 and x+6=y^2. This area can be computed

in two different ways using integrals

First of all it can be computed as a sum of two integrals as

integral f(x)dx from a,b + integral g(x)dx from b,c

I got a=-6
b=3
c=19

but what does
f(x)=?
g(x)?

I'm thinking it is sqrt(x+6) and (9-x)/2 but it tell me I am getting it wrong.

Ty for your help!

To find the area between the two graphs, we need to determine the points of intersection. Let's solve the system of equations formed by x+2y=9 and x+6=y^2.

1. Solve for x in terms of y in the first equation:
x = 9 - 2y

2. Substitute this value of x into the second equation to get:
9 - 2y + 6 = y^2

3. Rearrange the equation to get:
y^2 + 2y - 15 = 0

4. Factor the quadratic equation:
(y + 5)(y - 3) = 0

Therefore, y = -5 or y = 3.

Now, let's find the x-values at these y-values.

For y = -5:
x = 9 - 2(-5) = 19

For y = 3:
x = 9 - 2(3) = 3

So, the points of intersection are (19, -5) and (3, 3).

Now, let's determine the functions f(x) and g(x) to set up the integrals.

1. f(x) corresponds to the graph x+2y=9. Rearrange the equation to solve for y:
y = (9 - x)/2

2. g(x) corresponds to the graph x+6=y^2. Take the square root of both sides:
y = sqrt(x+6)

Therefore, f(x) = (9 - x)/2 and g(x) = sqrt(x+6).

To find the area between the graphs, integrate f(x) and g(x) with respect to x between suitable x-values.

∫[a,b]f(x)dx + ∫[b,c]g(x)dx

In this case, you have correctly determined the x-values:
a = 3
b = 19

However, there seems to be a mistake in your functions f(x) and g(x).

Instead, the correct functions for this problem are:

f(x) = sqrt(x+6) and g(x) = (9 - x)/2

So, to compute the area, you need to evaluate the following integral:

∫[3,19]sqrt(x+6)dx + ∫[19,c](9 - x)/2 dx

Evaluate these integrals individually and calculate their sum.

You appear to have found their intersection correct as

(3,3) and (19,-5)

USING VERTICAL SLICES:
y^2 = x+6
so y = ± √(x+6) = ± (x+6)^(1/2)

so for the part from -6 to 3
height = (x+6)^ - ( -(x+6)^(1/2)
= 2(x+6)^(1/2)
integral of that is (4/3)(x+6)^(3/2)
area of part 1
= (4/3)(9)^(3/2) - (0)
= 36
2nd part of the region:
heigh = (1/2)(9-x) - (-(x+6)^(1/2) )
= (1/2)( (9-x) + (x+6)^(1/2) )
integral of that is
(1/2) ( 9x - x^2/2 + (2/3)(x+6)^(3/2) )
evaluated from 3 to 19
= (1/2) [ 171 - 361/2 + (2/3)(125) - (27 - 9/2 + (2/3)(27) ]
= 50/3

better check my arithmetic, I was too lazy to write it out first.
So the area of the enclosed region using vertical slices = 36 + 50/3 = 158/3

USING HORIZONTAL SLICES:
width of a slice = 9-2y - (y^2 - 6)
= 15 - 2y - y^2
Area = ∫(15-2y-y^2) dy from -5 to 3
= [15y - y^2 - y^3/3] from -5 yo 3
= 45 - 9 - 27/3 - (-75 - 25 + 125/3)
= 256/3

ARGGG!! , error somewhere , sorry, perhaps you can spot it .