Solved on Feb 22, 2024

Find the area between the cubic curves $y=x^{3}-13x^{2}+30x$ and $y=-x^{3}+13x^{2}-30x$ .

STEP 1

Assumptions
1. We are given two functions: $y = x^3 - 13x^2 + 30x$ and $y = -x^3 + 13x^2 - 30x$ .
2. We need to find the area between these two curves.
3. The area between the curves is the integral of the upper function minus the integral of the lower function between the points of intersection.

STEP 2

First, we need to find the points of intersection between the two curves. This is where the two functions are equal to each other.
$x^3 - 13x^2 + 30x = -x^3 + 13x^2 - 30x$

STEP 3

Simplify the equation by moving all terms to one side.
$2x^3 - 26x^2 + 60x = 0$

STEP 4

Factor out the common term $x$ .
$x(2x^2 - 26x + 60) = 0$

STEP 5

Solve for $x$ by setting each factor equal to zero.
$x = 0 \quad \text{or} \quad 2x^2 - 26x + 60 = 0$

STEP 6

Factor the quadratic equation or use the quadratic formula to find the roots.
$2x^2 - 26x + 60 = (2x - 10)(x - 6) = 0$

STEP 7

Solve for $x$ .
$x = 5 \quad \text{or} \quad x = 6$

STEP 8

Now we have the points of intersection: $x = 0$ , $x = 5$ , and $x = 6$ .

STEP 9

To find the area between the curves, we need to integrate the difference between the two functions from $x = 0$ to $x = 5$ and from $x = 5$ to $x = 6$ .

STEP 10

Set up the integral for the area from $x = 0$ to $x = 5$ .
$Area_{1} = \int_{0}^{5} [(-x^3 + 13x^2 - 30x) - (x^3 - 13x^2 + 30x)] \, dx$

STEP 11

Simplify the integrand by combining like terms.
$Area_{1} = \int_{0}^{5} [-2x^3 + 26x^2 - 60x] \, dx$

STEP 12

Integrate the function with respect to $x$ .
$Area_{1} = \left[ -\frac{1}{2}x^4 + \frac{26}{3}x^3 - 30x^2 \right]_{0}^{5}$

STEP 13

Evaluate the integral at the bounds $x = 0$ and $x = 5$ .
$Area_{1} = \left( -\frac{1}{2}(5)^4 + \frac{26}{3}(5)^3 - 30(5)^2 \right) - \left( -\frac{1}{2}(0)^4 + \frac{26}{3}(0)^3 - 30(0)^2 \right)$

STEP 14

Calculate the value of the integral from $x = 0$ to $x = 5$ .
$Area_{1} = \left( -\frac{1}{2}(625) + \frac{26}{3}(125) - 30(25) \right) - (0)$
$Area_{1} = \left( -312.5 + \frac{3250}{3} - 750 \right)$
$Area_{1} = \left( -312.5 + 1083.33 - 750 \right)$
$Area_{1} = 20.83$

STEP 15

Set up the integral for the area from $x = 5$ to $x = 6$ .
$Area_{2} = \int_{5}^{6} [(x^3 - 13x^2 + 30x) - (-x^3 + 13x^2 - 30x)] \, dx$

STEP 16

Simplify the integrand by combining like terms.
$Area_{2} = \int_{5}^{6} [2x^3 - 26x^2 + 60x] \, dx$

STEP 17

Integrate the function with respect to $x$ .
$Area_{2} = \left[ \frac{1}{2}x^4 - \frac{26}{3}x^3 + 30x^2 \right]_{5}^{6}$

STEP 18

Evaluate the integral at the bounds $x = 5$ and $x = 6$ .
$Area_{2} = \left( \frac{1}{2}(6)^4 - \frac{26}{3}(6)^3 + 30(6)^2 \right) - \left( \frac{1}{2}(5)^4 - \frac{26}{3}(5)^3 + 30(5)^2 \right)$

STEP 19

Calculate the value of the integral from $x = 5$ to $x = 6$ .
$Area_{2} = \left( \frac{1}{2}(1296) - \frac{26}{3}(216) + 30(36) \right) - \left( \frac{1}{2}(625) - \frac{26}{3}(125) + 30(25) \right)$
$Area_{2} = \left( 648 - \frac{5616}{3} + 1080 \right) - \left( 312.5 - \frac{3250}{3} + 750 \right)$
$Area_{2} = \left( 648 - 1872 + 1080 \right) - \left( 312.5 - 1083.33 + 750 \right)$
$Area_{2} = -144 + 20.83$
$Area_{2} = -123.17$

STEP 20

Since area cannot be negative, we take the absolute value of $Area_{2}$ .
$Area_{2} = | -123.17 | = 123.17$

STEP 21

Add the two areas to find the total area between the curves.
$Total\, Area = Area_{1} + Area_{2}$
$Total\, Area = 20.83 + 123.17$

STEP 22

Calculate the total area.
$Total\, Area = 144$
The area between the curves is 144 square units.

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Find the area between the cubic curves y=x3−13x2+30xy=x^{3}-13x^{2}+30xy=x3−13x2+30x and y=−x3+13x2−30xy=-x^{3}+13x^{2}-30xy=−x3+13x2−30x.

STEP 1

STEP 2

First, we need to find the points of intersection between the two curves. This is where the two functions are equal to each other.x3−13x2+30x=−x3+13x2−30xx^3 - 13x^2 + 30x = -x^3 + 13x^2 - 30xx3−13x2+30x=−x3+13x2−30x

STEP 3

Simplify the equation by moving all terms to one side.2x3−26x2+60x=02x^3 - 26x^2 + 60x = 02x3−26x2+60x=0

STEP 4

Factor out the common term xxx.x(2x2−26x+60)=0x(2x^2 - 26x + 60) = 0x(2x2−26x+60)=0

STEP 5

Solve for xxx by setting each factor equal to zero.x=0or2x2−26x+60=0x = 0 \quad \text{or} \quad 2x^2 - 26x + 60 = 0x=0or2x2−26x+60=0

STEP 6

Factor the quadratic equation or use the quadratic formula to find the roots.2x2−26x+60=(2x−10)(x−6)=02x^2 - 26x + 60 = (2x - 10)(x - 6) = 02x2−26x+60=(2x−10)(x−6)=0

STEP 7

Solve for xxx.x=5orx=6x = 5 \quad \text{or} \quad x = 6x=5orx=6

STEP 8

Now we have the points of intersection: x=0x = 0x=0, x=5x = 5x=5, and x=6x = 6x=6.

STEP 9

To find the area between the curves, we need to integrate the difference between the two functions from x=0x = 0x=0 to x=5x = 5x=5 and from x=5x = 5x=5 to x=6x = 6x=6.

STEP 10

Set up the integral for the area from x=0x = 0x=0 to x=5x = 5x=5.Area1=∫05[(−x3+13x2−30x)−(x3−13x2+30x)] dxArea_{1} = \int_{0}^{5} [(-x^3 + 13x^2 - 30x) - (x^3 - 13x^2 + 30x)] \, dxArea1​=∫05​[(−x3+13x2−30x)−(x3−13x2+30x)]dx

STEP 11

Simplify the integrand by combining like terms.Area1=∫05[−2x3+26x2−60x] dxArea_{1} = \int_{0}^{5} [-2x^3 + 26x^2 - 60x] \, dxArea1​=∫05​[−2x3+26x2−60x]dx

STEP 12

Integrate the function with respect to xxx.Area1=[−12x4+263x3−30x2]05Area_{1} = \left[ -\frac{1}{2}x^4 + \frac{26}{3}x^3 - 30x^2 \right]_{0}^{5}Area1​=[−21​x4+326​x3−30x2]05​

STEP 13

STEP 14

STEP 15

Set up the integral for the area from x=5x = 5x=5 to x=6x = 6x=6.Area2=∫56[(x3−13x2+30x)−(−x3+13x2−30x)] dxArea_{2} = \int_{5}^{6} [(x^3 - 13x^2 + 30x) - (-x^3 + 13x^2 - 30x)] \, dxArea2​=∫56​[(x3−13x2+30x)−(−x3+13x2−30x)]dx

STEP 16

Simplify the integrand by combining like terms.Area2=∫56[2x3−26x2+60x] dxArea_{2} = \int_{5}^{6} [2x^3 - 26x^2 + 60x] \, dxArea2​=∫56​[2x3−26x2+60x]dx

STEP 17

Integrate the function with respect to xxx.Area2=[12x4−263x3+30x2]56Area_{2} = \left[ \frac{1}{2}x^4 - \frac{26}{3}x^3 + 30x^2 \right]_{5}^{6}Area2​=[21​x4−326​x3+30x2]56​

STEP 18

STEP 19

STEP 20

Since area cannot be negative, we take the absolute value of Area2Area_{2}Area2​.Area2=∣−123.17∣=123.17Area_{2} = | -123.17 | = 123.17Area2​=∣−123.17∣=123.17

STEP 21

Add the two areas to find the total area between the curves.Total Area=Area1+Area2Total\, Area = Area_{1} + Area_{2}TotalArea=Area1​+Area2​Total Area=20.83+123.17Total\, Area = 20.83 + 123.17TotalArea=20.83+123.17

STEP 22

Calculate the total area.Total Area=144Total\, Area = 144TotalArea=144The area between the curves is 144 square units.

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Find the area between the cubic curves $y=x^{3}-13x^{2}+30x$ and $y=-x^{3}+13x^{2}-30x$ .

First, we need to find the points of intersection between the two curves. This is where the two functions are equal to each other.
$x^3 - 13x^2 + 30x = -x^3 + 13x^2 - 30x$

Simplify the equation by moving all terms to one side.
$2x^3 - 26x^2 + 60x = 0$

Factor out the common term $x$ .
$x(2x^2 - 26x + 60) = 0$

Solve for $x$ by setting each factor equal to zero.
$x = 0 \quad \text{or} \quad 2x^2 - 26x + 60 = 0$

Factor the quadratic equation or use the quadratic formula to find the roots.
$2x^2 - 26x + 60 = (2x - 10)(x - 6) = 0$

Solve for $x$ .
$x = 5 \quad \text{or} \quad x = 6$

Now we have the points of intersection: $x = 0$ , $x = 5$ , and $x = 6$ .

To find the area between the curves, we need to integrate the difference between the two functions from $x = 0$ to $x = 5$ and from $x = 5$ to $x = 6$ .

Set up the integral for the area from $x = 0$ to $x = 5$ .
$Area_{1} = \int_{0}^{5} [(-x^3 + 13x^2 - 30x) - (x^3 - 13x^2 + 30x)] \, dx$

Simplify the integrand by combining like terms.
$Area_{1} = \int_{0}^{5} [-2x^3 + 26x^2 - 60x] \, dx$

Integrate the function with respect to $x$ .
$Area_{1} = \left[ -\frac{1}{2}x^4 + \frac{26}{3}x^3 - 30x^2 \right]_{0}^{5}$

Set up the integral for the area from $x = 5$ to $x = 6$ .
$Area_{2} = \int_{5}^{6} [(x^3 - 13x^2 + 30x) - (-x^3 + 13x^2 - 30x)] \, dx$

Simplify the integrand by combining like terms.
$Area_{2} = \int_{5}^{6} [2x^3 - 26x^2 + 60x] \, dx$

Integrate the function with respect to $x$ .
$Area_{2} = \left[ \frac{1}{2}x^4 - \frac{26}{3}x^3 + 30x^2 \right]_{5}^{6}$

Since area cannot be negative, we take the absolute value of $Area_{2}$ .
$Area_{2} = | -123.17 | = 123.17$

Add the two areas to find the total area between the curves.
$Total\, Area = Area_{1} + Area_{2}$
$Total\, Area = 20.83 + 123.17$

Calculate the total area.
$Total\, Area = 144$
The area between the curves is 144 square units.