Question

Sketch the following regions (if a figure is not given) and then find the area. The regions bounded by \(y=x^{2}(3-x)\) and \(y=12-4 x\)

Short answer

Answer: The area between the given curves is \(\frac{59}{4}\).

Step-by-step solution

Sketch the curves

First, we will sketch the curves \(y=x^2(3-x)\) and \(y=12-4x\) using a graphing calculator or software. Upon doing this, we can see that the region we need to find the area of is between these two curves.

Find the points of intersection

We need to find the points where the two curves intersect. To do this, we will set the equations equal to each other and solve for \(x\): $$x^2(3-x)=12-4x$$ $$x^2(3-x)+4x-12=0$$ This is a quadratic equation in terms of x. We can first distribute x into the parentheses: $$3x^2 - x^3 + 4x - 12=0$$ Now, we can rearrange the terms: $$x^3 - 3x^2 - 4x + 12 = 0$$ We can solve this cubic equation for \(x\) using either factoring, synthetic division or an appropriate calculator/software: The solutions are: \(x = 1,\ x = 2,\ x = 6\)

Set up and evaluate the definite integral

To find the area between the two curves, we can set up a definite integral that subtracts the lower curve from the upper curve. Our area formula will be: $$A = \int_{x_1}^{x_2} (upper\_curve - lower\_curve) dx$$ In this case, the points of intersection to be used are \(x=1\) and \(x=2\) because that's where the region is bounded: $$A = \int_{1}^{2} [(12-4x) - (x^2(3-x))] dx$$ Now, we will integrate with respect to \(x\): $$A = \int_{1}^{2}[-x^3 + 3x^2 - 4x +12]dx$$ Integrating: $$A = \left[-\frac{1}{4}x^4 + x^3 - 2 x^2 +12x\right]_{1}^{2}$$ Now we will plug in the limits of integration: $$A = \left(-\frac{1}{4}(2^4) + (2^3) - 2(2^2) + 12(2)\right) - \left(-\frac{1}{4}(1^4) + (1^3) - 2(1^2) + 12(1)\right)$$ Evaluating and simplifying: $$A = \left(-4 + 8 - 8 + 24\right) - \left(-\frac{1}{4} + 1 - 2 + 12\right)$$ $$A = 20 - \frac{21}{4} = \frac{59}{4}$$ Therefore, the area between the given curves is \(\frac{59}{4}\).

Key concepts

Cubic Equations

Cubic equations are mathematical expressions that involve a variable raised to the third power. In this problem, the equation was derived from setting two functions equal to each other. The original equation was simplified to a cubic form: \[x^3 - 3x^2 - 4x + 12 = 0\]Cubic equations like these can have up to three real roots. There are several methods to solve cubic equations:

• Factoring: If the equation can be factored easily, finding the roots is straightforward.
• Synthetic Division: This method is useful for simplifying the polynomial and finding potential roots.
• Numerical Methods or Calculators: For more complicated cubic equations, graphing calculators or specific software can help to pinpoint the roots.

In this example, the solution found the roots to be \(x = 1, x = 2, \text{and } x = 6\). These roots represent the x-values where the curves intersect.

Definite Integrals

Definite integrals are a central concept in calculus, providing a way to calculate areas under curves over a specific interval. The definite integral has the form:\[A = \int_{a}^{b} f(x) \, dx\]where \(a\) and \(b\) define the interval and \(f(x)\) is the function being integrated.In this exercise, the area between two curves is found by using a definite integral. You take the integral of the upper curve minus the lower curve. This looks like:\[A = \int_{1}^{2} [(12-4x) - (x^2(3-x))] dx\]This integration process involves finding the antiderivative, applying the bounds of the integration, and calculating the resulting expression to find the area. The solution shows these calculations, resulting in \[A = \frac{59}{4}\]which represents the area in question.

Intersection Points

Intersection points are where two or more graphs meet on a plane. Finding these points is essential for determining the boundaries of integration when calculating the area between curves. To find intersection points, you equate the functions and solve for the variable.In this exercise, the intersection points were found by solving the equation:\[x^2(3-x) = 12-4x\]After simplifying the equation, you could solve the derived cubic equation. The solutions \(x = 1, x = 2, \text{and } x = 6\) show where both given functions intersect on the x-axis. In the area calculation, only the range \(x = 1\) to \(x = 2\) is used, since this indicates where the region of interest lies between the curves within those bounds.

Sketching Graphs

Sketching graphs is fundamental to visualize problems involving areas between curves. The process starts with plotting the given functions. This exercise involves:

• The quadratic-cubic curve: \(y = x^2(3-x)\)
• The linear function: \(y = 12-4x\)

By plotting these graphs using graphing software or a calculator, we can visually understand the positions and relations of the curves and their intersection points. This visualization helps identify which function is the upper curve and which is the lower curve within the intersecting region. For areas between curves, it's crucial to know:

• Where the curves intersect (determines integration bounds).
• Which curve is on top and which is below (determines the function to subtract in integration).

By using sketches, you can effectively plan out solving steps and confirm calculations physically.