Consider the area between the graphs x+2y=4 and x+4=y2. This area can be computed

in two different ways using integrals

1) Compute as a sum of two integrals
2) Compute as a single integral
3) Either way, what is the area?

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So for 1) I rewrote the two to be:
x = 4-2y
x = y^2-4
and setting them equal:
y^2-4 = 4-2y gives:
y^2+2y-8 or (y+4)(y-2)
Therefore the roots are y = -4 and 2

So if its the integral of f(x) from a to b plus the integral of g(x) from b to c:
a: -4, b: 0, c: -12
But what would be the functions f(x) and g(x)? Nothing I tried has worked, and I have no idea how to approach it as a single integral.

Thanks!

p.s. I wrote y2 for the second equation but it is y^2

let's look at the graph

http://www.wolframalpha.com/input/?i=plot+x%2B2y%3D4+and+x%2B4%3Dy%5E2

You found the correct y values of the intersection points, the points are (0,2) and (12, -4)

So why not take horizontal slices?
the effective width = rightmost x - left x
= (4 - 2y) - (y^2 - 4)
= -y^2 - 2y + 8

area = ∫ (-y^2 - 2y + 8) dy from -4 to 2
= [ (-1/3)y^3 - y^2 + 8y] from -4 to 2
= ( (-1/3)(8) - 4 + 16) - ( (-1/3)(-64) - 16 - 32)
= -8/3 + 12 - 64/3 + 48
= 36

To compute the area between the graphs x+2y=4 and x+4=y^2 in two different ways using integrals, let's break it down step by step:

1) Compute as a sum of two integrals:
To find the area using two integrals, you need to split the region into two parts at the point of intersection (-4 and 2).

First, we need to determine the functions f(x) and g(x) for the intervals a to b and b to c:

For the interval a to b:
a: x = 4-2y
b: x = y^2-4

To find f(x), solve for y in terms of x:
4-2y = x
-2y = x - 4
y = (4-x)/2

For the interval b to c:
b: x = y^2-4
c: x = 0

Similarly, solving for g(x) in terms of y:
y^2 - 4 = x
y^2 = x + 4
y = √(x+4)

Now, calculate the integrals separately, summing them to find the total area:
Area = ∫[a,b] f(x) dx + ∫[b,c] g(x) dx

∫[a,b] f(x) dx = ∫[-4,0] (4-x)/2 dx
∫[a,b] f(x) dx = [(4x - x^2)/4] from -4 to 0

∫[b,c] g(x) dx = ∫[0,-12] √(x+4) dx
∫[b,c] g(x) dx = (2/3) * (√(x+4) * (x+4)^(3/2)) from 0 to -12

Finally, subtract the definite integrals from each other:

Area = [(4x - x^2)/4] from -4 to 0 - (2/3) * (√(x+4) * (x+4)^(3/2)) from 0 to -12

Evaluate each integral individually to find the area.

2) Compute as a single integral:
To compute the area as a single integral, we need to express the two curves in terms of x rather than y and rewrite the equation in terms of y:

x+2y = 4
x = 4 - 2y

x+4 = y^2
x = y^2 - 4

Setting these two equations equal to each other, we have:

4 - 2y = y^2 - 4
y^2 - 2y - 8 = 0

Solving this quadratic equation, we find y = -2 and y = 4.

The single integral for finding the area can be expressed as:

Area = ∫[a,b] (g(x) - f(x)) dx,

where a and b are the x-values of the points of intersection (-4 and 2).

Substituting the equations for f(x) and g(x) obtained earlier:

Area = ∫[-4,2] [(y^2 - 4) - (4 - 2y)] dy

Now, integrate the equation with respect to y over the interval [-4, 2] to find the area.

Whichever method you choose (sum of two integrals or single integral), the resulting value will give you the area between the two graphs.

To find the area between the graphs x+2y=4 and x+4=y^2, let's start by graphing these two equations.

To find the roots of y, you correctly set the two equations equal to each other and solved for y. The resulting quadratic equation is y^2 + 2y - 8 = 0, which can be factored as (y + 4)(y - 2) = 0. So, the roots are y = -4 and y = 2.

Now, let's find the x-values for these corresponding y-values. For y = -4, we substitute it into one of the equations to solve for x: x + 2(-4) = 4, which gives x = 12. Similarly, for y = 2, we have x + 2(2) = 4, which results in x = 0.

So, as you correctly identified, the limits of integration for the sum of two integrals approach are a = -4, b = 0, and c = 12.

Now, let's express the given area as a sum of two integrals:

1) Integrate with respect to y from -4 to 2:
Area_1 = ∫[y=-4 to 2] (x+4-y^2) dy

To evaluate this integral, we need to express x in terms of y. From the equation x + 2y = 4, we can solve it for x: x = 4 - 2y.

Substituting x = 4 - 2y into the integral expression, we get:
Area_1 = ∫[y=-4 to 2] (4 - 2y + 4 - y^2) dy
= ∫[y=-4 to 2] (8 - 2y - y^2) dy

Now, integrate the expression (8 - 2y - y^2) with respect to y and evaluate from y = -4 to y = 2 to calculate the area.

2) Compute as a single integral:

Now, let's consider the area as a single integral. We will integrate with respect to x from the values x = 0 to x = 12.

To rewrite the equations in terms of x, we have:
x = 4 - 2y => y = (4 - x) / 2
x + 4 = y^2 => y = √(x + 4)

So, for the single integral approach, the area can be calculated as:
Area_2 = ∫[x=0 to 12] (√(x+4) - (4 - x)/2) dx

Now, integrate the expression (√(x+4) - (4 - x)/2) with respect to x and evaluate from x = 0 to x = 12 to calculate the area.

To calculate the exact area values, you would need to evaluate the two integrals numerically using appropriate integration techniques or software.

The area calculated by both approaches should be the same, confirming the correctness of the calculation.