consider the area between the graphs x+3y=1 and x+9=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 where a=,b=,c= and f(x0= and g(x)=. Alterntaively this area can be computed as a single integral

Question

consider the area between the graphs x+3y=1 and x+9=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 where a=,b=,c= and f(x0= and g(x)=. Alterntaively this area can be computed as a single integral

Answer 1

this question is rallt hard

Answer 2

To compute the area between the two graphs, let's first find the points of intersection.

Given:
Equation 1: x + 3y = 1
Equation 2: x + 9 = y^2

To find the points of intersection, we need to solve these two equations simultaneously.

Step 1: Solve Equation 1 for x:
x = 1 - 3y

Step 2: Substitute x in Equation 2:
1 - 3y + 9 = y^2
10 = y^2 + 3y - 1

Step 3: Rearrange Equation 2 to a quadratic equation form:
y^2 + 3y - 11 = 0

Now we have a quadratic equation that we can solve using various methods, such as factoring or the quadratic formula.
However, in this case, it seems that the equation does not factor easily. Let’s use the quadratic formula:

Step 4: Apply the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Here a = 1, b = 3, and c = -11

Using these values in the formula, we get:
y = (-3 ± √(3^2 - 4(1)(-11))) / (2*1)
y = (-3 ± √(9 + 44)) / 2
y = (-3 ± √53) / 2

Now we have the two possible values of y at the points of intersection.

Step 5: Calculate the corresponding x-values:
Using the x = 1 - 3y equation, we can substitute the y-values found in Step 4 to find the corresponding x-values.

For the first y-value (y = (-3 + √53) / 2):
x = 1 - 3 * ((-3 + √53) / 2)

For the second y-value (y = (-3 - √53) / 2):
x = 1 - 3 * ((-3 - √53) / 2)

Now that we have the points of intersection, we can proceed to compute the area using two different methods: a sum of two integrals or a single integral.

Method 1: Sum of Two Integrals

Let's denote the points of intersection as (x0, y0) and (x1, y1), where x0 and x1 are the x-values we calculated in Step 5.

To compute the area using two integrals, we split the region into two parts vertically.

Integral 1: Calculate the area between Equation 1 and Equation 2 for the region from x = x0 to x = x1.
a = x0, b = x1, f(x) = x + 3y, g(x) = x + 9 - y^2

The area using Integral 1 can be calculated as:
Area1 = ∫[x0,x1] [g(x) - f(x)] dx

Integral 2: Calculate the area between Equation 2 and the x-axis for the region from x = x0 to x = x1.
a = x0, b = x1, f(x) = 0, g(x) = x + 9 - y^2

The area using Integral 2 can be calculated as:
Area2 = ∫[x0,x1] [g(x) - f(x)] dx

Finally, the total area between the two graphs is the sum of Area1 and Area2:
Total Area = Area1 + Area2

Method 2: Single Integral

Alternatively, we can compute the area using a single integral by integrating the absolute difference of the two functions.

To do this, we need to determine the function that lies above the other within the region. We can compare the values of the two functions at some point and figure out which is greater.

Let's assume that Equation 2 (x + 9 = y^2) is the upper function within the region.

Integral: Calculate the area between Equation 2 and Equation 1 for the region from x = x0 to x = x1.
a = x0, b = x1, f(x) = x + 3y, g(x) = x + 9 - y^2

The area using a single integral can be calculated as:
Total Area = ∫[x0,x1] |g(x) - f(x)| dx

Now you can substitute the values of x0 and x1 into the respective integrals or the single integral formula to compute the area between the two graphs.