Find the area of the region below y=x^2−2x+3 and above y=9 for 3≤x≤5.
Area =
Unclear. Below and above what?
below the parabola and above the line.
Unfortunately, the parabola is only above the line for x > 1+√7 = 3.6
Anyway, when you get things straightened out, the area is clearly just something like
∫[3.6,9] x^2-2x+3 - 9 dx
To find the area of the region below the curve y = x^2 - 2x + 3 and above the line y = 9, we need to find the area between these two curves within the given x-range.
Step 1: Find the points of intersection.
Set the two equations equal to each other: x^2 - 2x + 3 = 9.
Rearrange to get the quadratic equation: x^2 - 2x - 6 = 0.
Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, factoring gives: (x - 3)(x + 2) = 0.
So, x - 3 = 0 or x + 2 = 0.
This means x = 3 or x = -2.
Step 2: Evaluate the integrals.
To find the area between the curves between x = 3 and x = 5, we need to evaluate the integral of the top function minus the integral of the bottom function within this range.
Area = ∫[3,5] (x^2 - 2x + 3) dx - ∫[3,5] 9 dx.
Evaluate the first integral:
∫[3,5] (x^2 - 2x + 3) dx = [(1/3) x^3 - x^2 + 3x] | [3,5]
= [(1/3) (5^3) - (5^2) + 3(5)] - [(1/3) (3^3) - (3^2) + 3(3)]
= [125/3 - 25 + 15] - [27/3 - 9 + 9]
= [(125 - 75 + 45) - (27 - 9 + 9)]/3
= 95/3.
Evaluate the second integral:
∫[3,5] 9 dx = 9x | [3,5]
= 9(5) - 9(3)
= 45 - 27
= 18.
Step 3: Calculate the area.
Now subtract the second integral from the first:
Area = (95/3) - 18
= (95 - 54)/3
= 41/3.
Therefore, the area of the region below y = x^2 - 2x + 3 and above y = 9 for 3 ≤ x ≤ 5 is 41/3 square units.