Consider the area between the graphs x+y=16 and x+4= (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 integrate from a to b of f(x)dx + integrate from b to c of g(x)dx

What is the value of a, b, c and what are f(x) and g(x) equal to?

Alternatively this area can be computed as a single integral

integrate from alpha to beta of h(y)dy
Alpha=?, Beta=?, h(y)=?

Either way we find that the area is: ?

how did you get the -4 in your first step? i don't understand where that came from

To find the values of a, b, c, f(x), and g(x), we need to first evaluate the points of intersection between the graphs x+y=16 and x+4=y^2.

For the first method: sum of two integrals

1. Rewrite the equations to y in terms of x:
x + y = 16 => y = 16 - x
x + 4 = y^2 => y^2 = x + 4

2. Set the y expressions equal to each other and solve for x:
16 - x = y^2 => 16 - x = (x + 4)^2

Simplify the equation:
16 - x = (x^2 + 8x + 16)

Rearrange and combine like terms:
x^2 + 9x - 16 = 0

Solve the quadratic equation:
Using the quadratic formula, we have:
x = (-9 ± √(9^2 - 4(1)(-16))) / (2)
x = (-9 ± √(81 + 64)) / (2)
x = (-9 ± √145) / 2

The values of x are approximately -11.47 and 2.47.

3. Calculate the corresponding y values for each x value:
For x = -11.47, y = 16 - (-11.47) = 27.47
For x = 2.47, y = 16 - 2.47 = 13.53

So, our points of intersection are approximately (-11.47, 27.47) and (2.47, 13.53).

4. Now we can determine the values of a, b, c, f(x), and g(x):
a = -11.47
b = 2.47
c = 16

f(x) = y = 16 - x
g(x) = y^2 = x + 4

For the second method: single integral

1. Rewrite the equations to x in terms of y:
x + y = 16 => x = 16 - y
x + 4 = y^2 => x = y^2 - 4

2. Set the x expressions equal to each other and solve for y:
16 - y = y^2 - 4

Rearrange and combine like terms:
y^2 + y - 20 = 0

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

Solve for y:
y + 5 = 0 => y = -5
y - 4 = 0 => y = 4

The values of y are -5 and 4.

3. Calculate the corresponding x values for each y value:
For y = -5, x = 16 - (-5) = 21
For y = 4, x = 16 - 4 = 12

So, our points of intersection are (21, -5) and (12, 4).

4. Now we can determine the values of α, β, and h(y):
α = -5
β = 4
h(y) = x = 16 - y

To calculate the area between the graphs, we need to evaluate the integrals for both methods:

Method 1: Sum of two integrals
Area = ∫[a to b] f(x) dx + ∫[b to c] g(x) dx

Method 2: Single integral
Area = ∫[α to β] h(y) dy

Without specific bounds, it is not possible to provide a numerical value for the area.

To find the values of a, b, c, f(x), g(x), alpha, beta, and h(y), and ultimately the area between the given graphs, we need to first analyze the equations x+y=16 and x+4=y^2.

1. Sum of Two Integrals Approach:
For the first method, let's express the given equations in terms of y:

x + y = 16 ==> x = 16 - y (equation 1)
x + 4 = y^2 ==> x = y^2 - 4 (equation 2)

To determine the values of a, b, and c, we need to find the points of intersection between the graphs. Setting equation 1 equal to equation 2:

16 - y = y^2 - 4

Rearranging this quadratic equation:

y^2 + y - 20 = 0

Factoring or using the quadratic formula, we find that y = 4 or y = -5. Therefore, a = -5, b = 4, and c = 16.

Now, to find f(x) and g(x), integrate the respective functions over the given intervals:

f(x) = integrate(-5 to 4) [16 - x] dx
g(x) = integrate(4 to 16) [x + 4 - sqrt(x - 4)] dx

Evaluating these integrals will give you the values of f(x) and g(x).

2. Single Integral Approach:
For the second method, we need to express the given equations in terms of x:

x + y = 16 ==> y = 16 - x (equation 1)
x + 4 = y^2 ==> x = y^2 - 4 (equation 2)

To determine the values of alpha and beta, we need to find the y-values at the points of intersection between the graphs. Setting equation 1 equal to equation 2:

16 - x = x^2 - 4

Rearranging this quadratic equation:

x^2 + x - 20 = 0

Again, factoring or using the quadratic formula, we find that x = 4 or x = -5. Therefore, alpha = -5 and beta = 4.

To find h(y), we integrate the difference between the upper and lower x-values corresponding to the given y-interval:

h(y) = integrate(y^2 - 4 to 16) [16 - y] dy

Evaluating this integral will give you the value of h(y).

Once you have determined the values of a, b, c, f(x), g(x), alpha, beta, and h(y), you can proceed to calculate the area between the graphs using either method by evaluating the respective integrals.

Please let me know if you would like me to assist you in performing the integrations or if you have any further questions.

the graphs intersect at (12,4) and (21,-5)

a = ∫[-4,12] 2√(x+4) dx + ∫[12,21] 16-x+√(x+4) dx

a = ∫[-5,4] (16-y) - (y^2-4) dy

a = 243/2