Problem 19 - Calculus and its applications

Question

Find the area of the region bounded by the graphs of the given equations. $$ y=6 x-x^{2}, y=x $$

Step-by-Step Solution

Answer

The area of the region is \( \frac{125}{6} \).

Step 1: Find Points of Intersection

To determine where the two functions intersect, set the equations equal to each other: \( 6x - x^2 = x \). Simplify to find the points of intersection: \( -x^2 + 5x = 0 \). Factor this to get \( x(-x + 5) = 0 \). Therefore, the solutions are \( x = 0 \) and \( x = 5 \). Check these values by plugging back into the original equations: both give the points \((0,0)\) and \((5,5)\).

Step 2: Set Up the Integral for the Area

Determine which function is on top. For the interval \([0, 5]\), compare \( y = 6x - x^2 \) and \( y = x \). Since \( 6x - x^2 > x \) in this interval (except at the points), \( 6x - x^2 \) is the top curve. The integral is then the top function minus the bottom function over the interval: \[ \int_{0}^{5} ((6x - x^2) - x) \, dx. \]

Step 3: Simplify the Integral Expression

Simplify the integrand: \[\int_{0}^{5} (6x - x^2 - x) \, dx = \int_{0}^{5} (5x - x^2) \, dx. \]

Step 4: Calculate the Integral

Find the antiderivative of \( 5x - x^2 \): \[ \int (5x - x^2) \, dx = \frac{5}{2}x^2 - \frac{1}{3}x^3 + C. \] Evaluate this expression from \( x = 0 \) to \( x = 5 \): \[ \left[ \frac{5}{2}(5)^2 - \frac{1}{3}(5)^3 \right] - \left[ \frac{5}{2}(0)^2 - \frac{1}{3}(0)^3 \right]. \]

Step 5: Evaluate the Expression and Calculate the Area

Substitute into the expression: \[\left[ \frac{5}{2}(25) - \frac{1}{3}(125) \right] = \left[ \frac{125}{2} - \frac{125}{3} \right]. \] Find a common denominator and calculate: \[\frac{375}{6} - \frac{250}{6} = \frac{125}{6}. \] Thus, the area is \( \frac{125}{6} \).

Step 6: Simplify the Result (if necessary)

The simplest form for the area of the region is \( \frac{125}{6} \) which is approximately \( 20.833 \). However, \( \frac{125}{6} \) is the exact, simplest form of the area.

Key Concepts

Integration

Integration is one of the key tools in calculus that helps us calculate the area between two curves or under a curve. Imagine you want to find how much space lies between two overlapping functions; integration is what you'd use. In our exercise, we are looking at

• the polynomial function: \( y = 6x - x^2 \)
• and the linear function: \( y = x \)

To find the area between these curves, we first set up an integral with the top function minus the bottom function. This is crucial because it gives us the accumulated area as we move horizontally from one point of intersection to the next. For the interval \([0, 5]\), we found that \( 6x - x^2 \) sits above \( x \) as the top curve over this stretch. Integration then helps us determine the total space below \( 6x - x^2 - x \) from \( x = 0 \) to \( x = 5 \). Remember, when solving the definite integral, you'll find the antiderivative first and then substitute the limits to get a numerical value, representing the area.

Points of intersection

Finding the points where two graphs intersect is like asking, "Where do these two lines meet?" In our scenario, setting the two equations equal: \( 6x - x^2 = x \) helps locate these meeting spots. By simplifying: \( -x^2 + 5x = 0 \), we factor it into: \( x(-x + 5) = 0 \), giving us intersections at \( x = 0 \) and \( x = 5 \). This means our two curves meet at the points

• (0, 0)
• (5, 5).

This step establishes the boundaries of integration, as these solutions tell us how far to stretch our calculation to find the area between the curves. Without these points, we'd have no idea where our integral should start or end. Always check your calculations to make sure these points lie on both graphs. Often, plugging them back into the original equations verifies their legitimacy.

Polynomial functions

A polynomial function is composed of terms that are powers of a variable, each multiplied by a coefficient, like \( 6x - x^2 \) in our exercise. Here, we encounter a special type of polynomial known as a quadratic function because its highest exponent is 2. Quadratic polynomials graph as parabolas, which can open upwards or downwards. In this case, \( y = 6x - x^2 \) represents a downward-opening parabola.
Understanding polynomial functions help in predicting the "shape" of our area. While polynomials come in various forms, what’s important is how they influence integration. Quadratics like this serve as typical examples, with each part contributing to the overall area underneath when part of a bounded region calculation. Always observe the degree of the polynomial (the highest exponent) to anticipate its shape on a graph.