Problem 18 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=2 x^{3}-x^{4}, y=0$$

Step-by-Step Solution

Answer

The area bounded by the curves is \(\frac{8}{5}\).

Step 1: Identify the Intersection Points

To find the area between the curves, we need to determine where they intersect. Set the equation of the curve equal to zero: \[2x^3 - x^4 = 0\]Factor the equation to find the intersection points:\[x^3(2-x) = 0\]This gives us the solutions: \(x=0\) and \(x=2\). Hence, the curves intersect at these points.

Step 2: Set Up the Integral

The integral will give the area between \(y=2x^3 - x^4\) and \(y=0\) over the interval \([0, 2]\). Since it's between this function and the x-axis, the integral is straightforward:\[\int_{0}^{2}(2x^3 - x^4) \, dx\]

Step 3: Integrate the Function

Integrate the function \(2x^3 - x^4\) with respect to \(x\):\[\int(2x^3 - x^4) \, dx = \frac{2}{4}x^4 - \frac{1}{5}x^5 = \frac{1}{2}x^4 - \frac{1}{5}x^5\]

Step 4: Evaluate the Definite Integral

Evaluate the definite integral from 0 to 2:\[\left[\frac{1}{2}x^4 - \frac{1}{5}x^5\right]_0^2 = \left(\frac{1}{2}(2)^4 - \frac{1}{5}(2)^5\right) - \left(\frac{1}{2}(0)^4 - \frac{1}{5}(0)^5\right)\]\[= \left(\frac{1}{2} \times 16 - \frac{1}{5} \times 32\right)\]\[= 8 - \frac{32}{5}\]\[= \frac{40}{5} - \frac{32}{5}\]\[= \frac{8}{5}\]So, the area is \(\frac{8}{5}\).

Key Concepts

Definite Integral

A definite integral is a tool we use to calculate the area under a curve from one point to another along the x-axis. Imagine you have a curve drawn on a graph, and you want to find out how much space is beneath that curve in a certain interval. This is where the definite integral helps.

• It gives you a numerical value that represents an area.
• The calculation involves integrating the function between two specific points.

Here, in the problem, we are given the function \(y = 2x^3 - x^4\). We want to find the area under this curve from \(x=0\) to \(x=2\). The definite integral is written as \[ \int_{0}^{2} (2x^3 - x^4) \, dx \]. This notation means that we calculate the integral of the function \(2x^3 - x^4\) between 0 and 2. Once we solve this integral, it gives us a number that represents the area enclosed between the curve and the x-axis across the specified interval.

Intersection Points

Before calculating the area under the curves, it's important to know where they intersect. Intersection points are the x-values where different curves meet or cross each other. Finding these points is a crucial step because it tells us the limits of integration, the bounds within which we perform our calculations to find the area.
To identify the intersection points for the given curve \(y = 2x^3 - x^4\) and \(y=0\), we set the equation \(2x^3 - x^4 = 0\). We factored the equation to become \(x^3(2-x) = 0\), leading us to two solutions: \(x = 0\) and \(x = 2\). These solutions are the values of \(x\) where our curve intersects the x-axis, meaning that the area we wish to find is confined between these two points. This gives us our limits of integration, i.e., 0 and 2.

Area Under Curves

The area under a curve represents a physical quantity, such as the accumulated total or net change, obtained from integrating the function. Think of it as measuring the space between the graph of the function and the x-axis.
Finding the area under a curve requires using integration when the function is continuous over the interval of interest.
For the exercise given, we consider the function \(2x^3 - x^4\) between \(x=0\) and \(x=2\). Integrating this function tells us how much area it covers between these two points on the x-axis. You could visualize this as drawing the curve on graph paper and estimating the colored region's size underneath the curve between the specified x-values.
When computing the definite integral \[\int_{0}^{2} (2x^3 - x^4) \, dx\], the solution involves evaluating the integrated expression \[\left(\frac{1}{2}x^4 - \frac{1}{5}x^5\right)\]. Plugging in the limits (from 0 to 2) into this results in \[\left(\frac{1}{2} \times 16 - \frac{1}{5} \times 32\right)\], yielding the area as \(\frac{8}{5}\). This result of \(\frac{8}{5}\) represents the entire area covered by the curve and x-axis between the intersection points under the graph.