What is the area of the region located between the graphs of y = -x² + x + 4 and y = 2?
Thank you so much!
find the intersection of the two boundary functions
2 = - x^2 + x + 4
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
height of the region in that domain
= -x^2 + x + 4 -2
= -x^2 + x + 2
area = [integral](-x^2 + x + 2) form -1 to 2
= -(1/3)x^3 + (1/2)x^2 + 2x form -1 to 2
= (-8/3 + 2 + 4) - (1/3 + 1/2 - 2)
= .....
To find the area of the region located between the two graphs, you would need to find the points of intersection first. The region is bounded by the x-values where the two functions intersect.
Setting the two functions equal to each other, we have:
-x² + x + 4 = 2
Rearranging the equation, we get:
-x² + x + 2 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b² - 4ac))/(2a)
For this equation, a = -1, b = 1, and c = 2. Plugging these values into the quadratic formula, we get:
x = (-(1) ± √((1)² - 4(-1)(2)))/(2(-1))
Simplifying the equation further:
x = (-1 ± √(1 + 8))/(-2)
x = (-1 ± √9)/(-2)
Now, we have two possible solutions for x:
x₁ = (-1 + 3)/(-2) = -1
x₂ = (-1 - 3)/(-2) = 2
The points of intersection are x = -1 and x = 2. To find the y-values, we substitute these x-values into either of the original equations.
For y = -x² + x + 4, substituting x = -1 gives:
y = -(-1)² + (-1) + 4
y = -1 + (-1) + 4
y = 2
For y = 2, substituting x = 2 gives:
2 = -x² + x + 4
0 = -x² + x + 2
Now, we have the two points of intersection: (-1, 2) and (2, 2).
To find the area between the two curves, we integrate the difference of the two functions with respect to x, within the range from -1 to 2:
Area = ∫[a,b] (y₁ - y₂) dx
Using the given equations, the area is:
Area = ∫[-1,2] (-x² + x + 4 - 2) dx
Integrating:
Area = ∫[-1,2] (-x² + x + 2) dx
Evaluating the integral:
Area = [-(x³/3) + (x²/2) + 2x] [-1,2]
Plugging in the upper and lower limits:
Area = [-(2³/3) + (2²/2) + 2(2)] - [(-(-1)³/3) + ((-1)²/2) + 2(-1)]
Calculating:
Area = [-(8/3) + 2 + 4] - [(1/3) + (1/2) - 2]
Area = [-8/3 + 10/3 + 12/3] - [1/3 + 1/2 - 2]
Area = [14/3] - [1/3 + 3/6 - 2]
Area = [14/3] - [1/3 + 1/2 - 6/3]
Area = [14/3] - [1/3 + 2/6 - 6/3]
Area = [14/3] - [1/3 + 1/3 - 6/3]
Area = [14/3] - [-6/3]
Area = 20/3
So, the area of the region located between the graphs is 20/3 square units.