Question:
Find the area enclosed by the curves f(x) = x^2 - 2x and g(x) = x + 1.
Answer:
To find the area enclosed by the curves, we first need to determine the points of intersection between the two curves. We can do this by setting the equations equal to each other and solving for x:
x^2 - 2x = x + 1
Moving all terms to one side of the equation, we get:
x^2 - 3x - 1 = 0
Next, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -3, and c = -1. Plugging these values into the quadratic formula, we obtain:
x = (-(-3) ± √((-3)^2 - 4(1)(-1))) / (2(1)) x = (3 ± √(9 + 4)) / 2 x = (3 ± √13) / 2
So, the points of intersection of the curves are at x = (3 + √13) / 2 and x = (3 - √13) / 2.
The area enclosed by the curves is given by the definite integral:
A = ∫[a, b] (f(x) - g(x)) dx
Where a and b are the x-coordinates of the points of intersection. We need to take the absolute value of the difference between the two functions to avoid negative area.
Thus, our integral becomes:
A = ∫[(3 - √13)/2, (3 + √13)/2] |(x^2 - 2x) - (x + 1)| dx
Simplifying the integrand:
A = ∫[(3 - √13)/2, (3 + √13)/2] |x^2 - 3x - 1| dx
We can split the integral into two separate integrals, each corresponding to the areas between the curves in the intervals [a, c] and [c, b] where c is the x-coordinate of the point of intersection:
A = ∫[(3 - √13)/2, c] (3x - x^2 - 1) dx + ∫[c, (3 + √13)/2] (x^2 - 3x - 1) dx
Evaluating the integrals gives us:
A = [x^2/2 - (x^3)/3 - x] [(3 - √13)/2, c] + [(x^3)/3 - (3x^2)/2 - x] [c, (3 + √13)/2]
Substituting the values of x = (3 - √13)/2 and x = (3 + √13)/2 into above, we can find the total area.
A = [ (27 + 3√13)/2 - (9 - √13)/6 - (3 - √13)/2 - ( 3(3 - √13)/2 - ((3 - √13)/2)^3/3 - (3 - √13)/2 ]
After solving further, we get the final area value.
A = (√13 + 6) / 2 square units
So, the area enclosed by the curves is (√13 + 6) / 2 square units.