Problem 22 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the total area between the curves \(y=x^2(3-x)\) and \(y=12-4x\). Answer: The total area between the curves is \(\frac{63}{4}\) square units.

Step 1: Sketch the Graphs

Begin by sketching the graphs of each equation to visually see the region we are trying to find the area of. Graph \(y=x^2(3-x)\) and \(y=12-4x\) using a graphing calculator or software tool.

Step 2: Find the Points of Intersection

Now we need to find the points where the two curves intersect. To do this, we set them equal to each other and solve for \(x\) : \(x^2(3-x)=12-4x\) Rearrange the equation: \(x^3-x^2+4x-12=0\) Use factoring or a software tool to solve the equation, and we find the points of intersection: \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\) Now, find the corresponding \(y\) values by plugging those \(x\) values back into either equation (let's choose the second, simpler one, \(y=12-4x\)): \(y_1 = 12 - 4(1) = 8\) \(y_2 = 12 - 4(2) = 4\) \(y_3 = 12 - 4(3) = 0\) So, the points of intersections are \((1, 8)\), \((2, 4)\), and \((3, 0)\).

Step 3: Set Up the Integral

We will now set up an integral to find the area between the curves. Since we know the points of intersections, we can use these as our boundaries for the integral. The difference between the functions will give the height of the area strip. So, the integral to find the total area between the curves is: \(\int_{1}^{3}[12-4x - x^2(3-x)]dx\)

Step 4: Evaluate the Integral

Finally, we evaluate the integral to find the total area: \(\int_{1}^{3}[12-4x - x^2(3-x)]dx = \int_{1}^{3}(12-4x-3x^2+x^3)dx\) Perform the integration: \(F(x) = 12x - 2x^2 - x^3 + \frac{1}{4}x^4 + C \) Now, evaluate the definite integral using the bounds: \(F(3) - F(1) = \left[12(3) - 2(3)^2 - (3)^3 + \frac{1}{4}(3)^4\right] - \left[12(1) - 2(1)^2 - (1)^3 + \frac{1}{4}(1)^4\right]\) Calculate the result, and we find the total area between the curves: \(= 27−18−9+\frac{81}{4} - (12-2-1+\frac{1}{4})\) \(= 3 + \frac{51}{4}\) \(= \frac{63}{4}\) So, the total area between the curves is \(\frac{63}{4}\) square units.

Key Concepts

Definite Integral

Understanding the definite integral is crucial for calculating the area between curves in calculus. The definite integral represents the accumulation of quantities and can be visualized as the area under a curve on a graph. More formally, if you have a continuous function f(x) over an interval [a, b], the definite integral of f(x) from a to b is the net area between the x-axis and the curve from x=a to x=b.

When we encounter an exercise requiring us to find the area between two curves, like in our example, we subtract one function from the other to find the 'net' function which represents the height of the area at any given point. We then integrate this net function within the bounds determined by the intersection points of the two curves. The result from evaluating this definite integral gives us the total area between the curves, considering that one function may be above the other over different intervals.

Intersection of Curves

Identifying the points where two curves intersect is a foundational step in solving many calculus problems. In the context of finding the area between curves, determining these intersection points is essential as they define the limits of integration for the definite integral.

To find the points of intersection, we equate the two functions and solve for the common values of x. For the given exercise, setting the two equations equal to each other and solving, provides us with the exact x coordinates of the intersection points. We must also compute the corresponding y values to get the complete (x, y) coordinates. These intersection points are crucial as they delineate precisely where we start and stop calculating the area between the curves.

Graphing Functions

Graphing the functions involved in the problem is not only a step in solving the problem, but also a visual aid that can deepen our understanding of the relationship between the curves. It helps us see which function lies on top or below at different intervals and provides a visual confirmation of the intersection points we calculated algebraically.

In our exercise, graphing the quadratic function y = x^2(3-x) and the linear function y = 12-4x allows us to visualize the distinct areas we're interested in. The visual aspect of the graph can also serve as a check against possible algebraic errors when determining intersection points. Nowadays, there are many graphing calculators and software tools available that make this step more accessible and accurate.