Consider the area between the graphs x+1y=12 and x+8=y2 . This area can be computed in two different ways using integrals

First of all it can be computed as a sum of two integrals
Interval a to b
f(x)= i put 12-x but it's wrong
Interval c to b
f(x)= i put (x-8)^(1/2) but it's wrong as well
I graphed this problem and got a=-8 and b=8. But the number my graph shows for c is not right.

Part two is
Alternatively this area can be computed as a single integral
Interval a to b
h(y)= ?

I assumed that the endpoints would be the same as above but it's not right. Can someone explain the steps i can take to get it. I've been working on this for a couple of hours.

To find the area between the two graphs using integrals, let's break it down into two separate integrals.

First, let's find the interval (a to b) for the x-values where the two curves intersect. We have the equations:

1) x + y = 12
2) x + 8 = y^2

To find the intersecting points, we can solve these equations simultaneously:

From equation 1, we get: y = 12 - x
Substituting this in equation 2, we have: x + 8 = (12 - x)^2

Expanding and solving for x, we obtain: x^2 - 23x + 64 = 0
Factoring, we get: (x - 8)(x - 8) = 0

So, x = 8 is the only solution. Thus, a = -8 and b = 8.

Now, let's calculate the area between the graphs using two separate integrals.

For the first graph (x + y = 12), we need to express y in terms of x:
y = 12 - x.

The integral for this graph would be:
A1 = ∫[a to b] (12 - x) dx.

Integrating, we obtain:
A1 = [(12x - (1/2)x^2)] [a to b].

Evaluating the integral:
A1 = [(12b - (1/2)b^2)] - [(12a - (1/2)a^2)].

Now, for the second graph (x + 8 = y^2), we need to express x in terms of y:
x = y^2 - 8.

The integral for this graph would be:
A2 = ∫[c to b] (y^2 - 8) dy.

To find the values for c, we can substitute x = a = -8 in equation 2: -8 + 8 = y^2.
This gives us y = 0 as the lower limit of integration.

Therefore, A2 = ∫[0 to b] (y^2 - 8) dy.

Integrating, we obtain:
A2 = [(1/3)y^3 - 8y] [0 to b].

Evaluating the integral:
A2 = [(1/3)b^3 - 8b].

Now, let's move on to the alternative method of calculating the area using a single integral.

We need to express x in terms of y for both equations.

For the first graph (x + y = 12):
x = 12 - y.

For the second graph (x + 8 = y^2):
x = y^2 - 8.

Now, to find the interval for the y-values, we need to find the y-coordinate of the intersection point. From before, we found that x = 8 at the intersection point. So, we substitute x = 8 into equation 1:
8 + y = 12,
which gives us y = 4.

Therefore, the interval for the y-values is [0 to 4].

The single integral for the area is given by:
A3 = ∫[0 to 4] [(12 - y) - (y^2 - 8)] dy.

Simplifying, we have:
A3 = ∫[0 to 4] (20 - y - y^2) dy.

Integrating, we obtain:
A3 = [20y - (1/2)y^2 - (1/3)y^3] [0 to 4].

Evaluating the integral:
A3 = [80 - 32/2 - 64/3] - [0 - 0 + 0] = 28 + 32/3.

Thus, the area between the two graphs can be calculated using two separate integrals (A1 + A2) or a single integral (A3).

To find the area between the graphs x+1y=12 and x+8=y^2, you can approach it in two different ways using integrals.

First, let's consider the sum of two integrals. The idea is to find the intervals on the x-axis (a to b) where the two curves intersect and split the area into two parts.

Step 1: Set the two equations equal to each other to find the points of intersection:
x + 1y = 12
x + 8 = y^2

Set them equal: 12 = y^2 + 1y
Rearrange and set the equation equal to zero: y^2 + 1y - 12 = 0
Factor the equation: (y + 4)(y - 3) = 0
Therefore, y = -4 or y = 3

Step 2: Determine the x-values at these intersection points by substituting the y-values back into either equation:
For y = -4:
x + 1(-4) = 12
x - 4 = 12
x = 16

For y = 3:
x + 1(3) = 12
x + 3 = 12
x = 9

So, the points of intersection are (16, -4) and (9, 3).

Step 3: Set up the integrals to find the individual areas:

Interval a to b: (from x = 9 to x = 16)
For the upper curve, y = y^2 + 8
For the lower curve, y = 12 - x

So, the first integral would be ∫[9, 16] [y^2 + 8 - (12 - x)] dx.

Interval c to b: (from y = -4 to y = 3)
For the upper curve, y = y^2 + 8
For the lower curve, y = 12 - x

So, the second integral would be ∫[-4, 3] [(12 - y^2 - 8) - (12 - y^2)] dy.

Now, let's move on to the second approach of computing the area using a single integral.

Step 1: First, rearrange the equations in terms of x:
x = 12 - y
x = y^2 - 8

Step 2: Solve for y in terms of x:
12 - y = y^2 - 8
Rearrange the equation: y^2 + y - 20 = 0
Factor the equation: (y - 4)(y + 5) = 0
Therefore, y = 4 or y = -5

Step 3: Determine the y-values at these intersection points by substituting the x-values back into either equation:
For x = 12 - y:
x = 12 - 4
x = 8

For x = y^2 - 8:
x = (4)^2 - 8
x = 8

So, the points of intersection are (8, 4) and (8, -5).

Step 4: Set up the integral to find the area:

Interval a to b: (from y = -5 to y = 4)
For the curve, x = 12 - y

So, the integral would be ∫[-5, 4] (12 - y) dy.

By following these steps, you should be able to accurately compute the area between the two graphs using integrals.

You have a parabola opening to the right, crossed by a line. The two curves intersect at (8,4) and (17,-5).

It's much easier to integrate on y, so you end up with

∫[-5,4] (12-y)-(y^2-8) dy = 243/2