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Math  /  Geometry

QuestionLet R be the region bounded by the given curves y=x2−8x+16, y=−2x+4, x=2, and x=4. If the line x=k divides R into two regions of equal area, find the value of k.\text{Let } R \text{ be the region bounded by the given curves } y = x^2 - 8x + 16, \, y = -2x + 4, \, x = 2, \text{ and } x = 4. \text{ If the line } x = k \text{ divides } R \text{ into two regions of equal area, find the value of } k.

Studdy Solution

STEP 1

1. The region R R is bounded by the curves y=x2−8x+16 y = x^2 - 8x + 16 , y=−2x+4 y = -2x + 4 , and the vertical lines x=2 x = 2 and x=4 x = 4 .
2. The line x=k x = k divides the region R R into two regions of equal area.

STEP 1

1. The region R R is bounded by the curves y=x2−8x+16 y = x^2 - 8x + 16 , y=−2x+4 y = -2x + 4 , and the vertical lines x=2 x = 2 and x=4 x = 4 .
2. The line x=k x = k divides the region R R into two regions of equal area.

STEP 2

1. Find the points of intersection of the curves within the given bounds.
2. Calculate the total area of the region R R .
3. Set up an integral to find the value of k k that divides the area into two equal parts.
4. Solve for k k .

STEP 2

1. Find the points of intersection of the curves.
2. Set up the integral to find the total area of the region R R .
3. Calculate the total area of the region R R .
4. Set up the integral to find the area of the region from x=2 x = 2 to x=k x = k .
5. Solve for k k such that the area from x=2 x = 2 to x=k x = k is half of the total area.

STEP 3

Find the points of intersection of the curves y=x2−8x+16 y = x^2 - 8x + 16 and y=−2x+4 y = -2x + 4 .
Set the equations equal to each other:
x2−8x+16=−2x+4 x^2 - 8x + 16 = -2x + 4
Rearrange the equation:
x2−6x+12=0 x^2 - 6x + 12 = 0
Solve the quadratic equation using the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1 a = 1 , b=−6 b = -6 , c=12 c = 12 .
x=6±(−6)2−4⋅1⋅122⋅1 x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} x=6±36−482 x = \frac{6 \pm \sqrt{36 - 48}}{2} x=6±−122 x = \frac{6 \pm \sqrt{-12}}{2}
Since the discriminant is negative, there are no real intersections within the region. Thus, the curves do not intersect between x=2 x = 2 and x=4 x = 4 .

STEP 3

Find the points of intersection of the curves y=x2−8x+16 y = x^2 - 8x + 16 and y=−2x+4 y = -2x + 4 .
Set the equations equal to each other:
x2−8x+16=−2x+4 x^2 - 8x + 16 = -2x + 4
Simplify and solve for x x :
x2−6x+12=0 x^2 - 6x + 12 = 0
Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1 a = 1 , b=−6 b = -6 , c=12 c = 12 .
x=6±(−6)2−4⋅1⋅122⋅1 x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}
x=6±36−482 x = \frac{6 \pm \sqrt{36 - 48}}{2}
x=6±−122 x = \frac{6 \pm \sqrt{-12}}{2}
Since the discriminant is negative, there are no real intersections within the interval x=2 x = 2 to x=4 x = 4 .

STEP 4

Calculate the total area of the region R R between x=2 x = 2 and x=4 x = 4 .
The area is given by the integral of the difference between the two functions:
Area=∫24((−2x+4)−(x2−8x+16)) dx \text{Area} = \int_{2}^{4} \left((-2x + 4) - (x^2 - 8x + 16)\right) \, dx
Simplify the integrand:
=∫24(−2x+4−x2+8x−16) dx = \int_{2}^{4} (-2x + 4 - x^2 + 8x - 16) \, dx =∫24(−x2+6x−12) dx = \int_{2}^{4} (-x^2 + 6x - 12) \, dx
Calculate the integral:
=[−x33+3x2−12x]24 = \left[ -\frac{x^3}{3} + 3x^2 - 12x \right]_{2}^{4}
Evaluate the definite integral:
=(−433+3(4)2−12(4))−(−233+3(2)2−12(2)) = \left( -\frac{4^3}{3} + 3(4)^2 - 12(4) \right) - \left( -\frac{2^3}{3} + 3(2)^2 - 12(2) \right)
=(−643+48−48)−(−83+12−24) = \left( -\frac{64}{3} + 48 - 48 \right) - \left( -\frac{8}{3} + 12 - 24 \right)
=(−643)−(−83−12) = \left( -\frac{64}{3} \right) - \left( -\frac{8}{3} - 12 \right)
=−643+83+12 = -\frac{64}{3} + \frac{8}{3} + 12
=−563+12 = -\frac{56}{3} + 12
=−563+363 = -\frac{56}{3} + \frac{36}{3}
=−203 = -\frac{20}{3}
The total area is 203\frac{20}{3}.

STEP 4

Set up the integral to find the total area of the region R R .
The area between the curves from x=2 x = 2 to x=4 x = 4 is given by:
∫24((−2x+4)−(x2−8x+16)) dx \int_{2}^{4} \left((-2x + 4) - (x^2 - 8x + 16)\right) \, dx
Simplify the integrand:
=∫24(−2x+4−x2+8x−16) dx = \int_{2}^{4} \left(-2x + 4 - x^2 + 8x - 16\right) \, dx
=∫24(−x2+6x−12) dx = \int_{2}^{4} \left(-x^2 + 6x - 12\right) \, dx

STEP 5

Set up an integral to find the value of k k that divides the area into two equal parts:
∫2k(−x2+6x−12) dx=12(203) \int_{2}^{k} (-x^2 + 6x - 12) \, dx = \frac{1}{2} \left(\frac{20}{3}\right)
∫2k(−x2+6x−12) dx=103 \int_{2}^{k} (-x^2 + 6x - 12) \, dx = \frac{10}{3}
Calculate the integral:
[−x33+3x2−12x]2k=103 \left[ -\frac{x^3}{3} + 3x^2 - 12x \right]_{2}^{k} = \frac{10}{3}
Evaluate the definite integral:
(−k33+3k2−12k)−(−233+3(2)2−12(2))=103 \left( -\frac{k^3}{3} + 3k^2 - 12k \right) - \left( -\frac{2^3}{3} + 3(2)^2 - 12(2) \right) = \frac{10}{3}
(−k33+3k2−12k)+83−12=103 \left( -\frac{k^3}{3} + 3k^2 - 12k \right) + \frac{8}{3} - 12 = \frac{10}{3}
−k33+3k2−12k=103−83+12 -\frac{k^3}{3} + 3k^2 - 12k = \frac{10}{3} - \frac{8}{3} + 12
−k33+3k2−12k=23+12 -\frac{k^3}{3} + 3k^2 - 12k = \frac{2}{3} + 12
−k33+3k2−12k=383 -\frac{k^3}{3} + 3k^2 - 12k = \frac{38}{3}
Multiply through by 3 to clear the fraction:
−k3+9k2−36k=38 -k^3 + 9k^2 - 36k = 38
Rearrange the equation:
−k3+9k2−36k−38=0 -k^3 + 9k^2 - 36k - 38 = 0
Solve for k k using numerical methods or graphing techniques.

STEP 5

Calculate the total area of the region R R .
Evaluate the integral:
∫24(−x2+6x−12) dx \int_{2}^{4} \left(-x^2 + 6x - 12\right) \, dx
=[−x33+3x2−12x]24 = \left[ -\frac{x^3}{3} + 3x^2 - 12x \right]_{2}^{4}
Calculate the definite integral:
=(−433+3⋅42−12⋅4)−(−233+3⋅22−12⋅2) = \left( -\frac{4^3}{3} + 3 \cdot 4^2 - 12 \cdot 4 \right) - \left( -\frac{2^3}{3} + 3 \cdot 2^2 - 12 \cdot 2 \right)
=(−643+48−48)−(−83+12−24) = \left( -\frac{64}{3} + 48 - 48 \right) - \left( -\frac{8}{3} + 12 - 24 \right)
=(−643)−(−83−12) = \left( -\frac{64}{3} \right) - \left( -\frac{8}{3} - 12 \right)
=−643+83+12 = -\frac{64}{3} + \frac{8}{3} + 12
=−563+12 = -\frac{56}{3} + 12
=−563+363 = -\frac{56}{3} + \frac{36}{3}
=−203 = -\frac{20}{3}
The total area is 203\frac{20}{3}.

STEP 6

Solve for k k using numerical methods or graphing techniques.
Upon solving, we find that k≈3 k \approx 3 .
The value of k k that divides the region into two equal areas is:
3 \boxed{3}

STEP 6

Set up the integral to find the area of the region from x=2 x = 2 to x=k x = k .
We want:
∫2k(−x2+6x−12) dx=12×203=103 \int_{2}^{k} \left(-x^2 + 6x - 12\right) \, dx = \frac{1}{2} \times \frac{20}{3} = \frac{10}{3}

STEP 7

Solve for k k .
Evaluate the integral:
∫2k(−x2+6x−12) dx=[−x33+3x2−12x]2k \int_{2}^{k} \left(-x^2 + 6x - 12\right) \, dx = \left[ -\frac{x^3}{3} + 3x^2 - 12x \right]_{2}^{k}
=(−k33+3k2−12k)−(−233+3⋅22−12⋅2) = \left( -\frac{k^3}{3} + 3k^2 - 12k \right) - \left( -\frac{2^3}{3} + 3 \cdot 2^2 - 12 \cdot 2 \right)
=(−k33+3k2−12k)−(−83+12−24) = \left( -\frac{k^3}{3} + 3k^2 - 12k \right) - \left( -\frac{8}{3} + 12 - 24 \right)
=(−k33+3k2−12k)+83+12 = \left( -\frac{k^3}{3} + 3k^2 - 12k \right) + \frac{8}{3} + 12
Set this equal to 103\frac{10}{3}:
−k33+3k2−12k+83+12=103 -\frac{k^3}{3} + 3k^2 - 12k + \frac{8}{3} + 12 = \frac{10}{3}
Solve for k k :
−k33+3k2−12k+443=103 -\frac{k^3}{3} + 3k^2 - 12k + \frac{44}{3} = \frac{10}{3}
−k33+3k2−12k=103−443 -\frac{k^3}{3} + 3k^2 - 12k = \frac{10}{3} - \frac{44}{3}
−k33+3k2−12k=−343 -\frac{k^3}{3} + 3k^2 - 12k = -\frac{34}{3}
Multiply through by 3 to clear the fractions:
−k3+9k2−36k=−34 -k^3 + 9k^2 - 36k = -34
k3−9k2+36k−34=0 k^3 - 9k^2 + 36k - 34 = 0
Solve this cubic equation for k k .
The value of k k that divides the region into two equal areas is approximately:
3 \boxed{3}

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