Given the functions f(x) = x + 1 and g(x) = (x + 1)^3
Find the area of the region enclosed by the graphs of f and g
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To find the area of the region enclosed by the graphs of f and g, we need to find the points of intersection between the two graphs and then integrate the difference between the two functions over the interval between these points.
Setting f(x) equal to g(x), we have:
x + 1 = (x + 1)^3
Expanding (x + 1)^3, we get:
x + 1 = x^3 + 3x^2 + 3x + 1
Rearranging and simplifying, we have:
x^3 + 3x^2 + 3x = x
Combining like terms, we get:
x^3 + 3x^2 + 2x = 0
Factoring out an x, we get:
x(x^2 + 3x + 2) = 0
Setting each factor equal to zero, we have:
x = 0, x^2 + 3x + 2 = 0
Solving the quadratic equation x^2 + 3x + 2 = 0, we get:
(x + 1)(x + 2) = 0
x = -1, -2
Therefore, the points of intersection between the two graphs are (-1, 0) and (-2, 1).
To find the area of the region enclosed by the graphs of f and g, we integrate the difference between the two functions over the interval [-2, -1].
The area is given by:
A = ∫[from -2 to -1] (g(x) - f(x)) dx
= ∫[from -2 to -1] ((x + 1)^3 - (x + 1)) dx
Expanding and simplifying the integrand, we have:
A = ∫[from -2 to -1] (x^3 + 3x^2 + 3x) dx
Integrating each term separately, we have:
A = [¼x^4 + x^3 + (3/2)x^2] from -2 to -1
= [(¼(-1)^4 + (-1)^3 + (3/2)(-1)^2) - (¼(-2)^4 + (-2)^3 + (3/2)(-2)^2)]
Evaluating the expression, we get:
A = [(¼ + (-1) + (3/2)) - (¼(16) + (-8) + (3/2)(4))]
= [¼ + (-1) + (3/2) - 4 + 8 - 6]
= [-¾]
Therefore, the area of the region enclosed by the graphs of f and g is -¾ square units.