Problem 18 - Calculus Early Transcendentals

Question

Sketch each region (if a figure is not given) and then find its total area. The regions in the first quadrant on the interval [0,2] bounded by \(y=4 x-x^{2}\) and \(y=4 x-4\)

Step-by-Step Solution

Answer

Answer: The total area of the region in the first quadrant bounded by the given curves on the interval [0,2] is \(\frac{16}{3}\).

Step 1: Sketch the curves

In order to sketch, first we need to know what the given functions look like. The two given functions are \(y = 4x - x^2\) and \(y = 4x - 4\) Both functions are linear in \(x\) and belong to a family of quadratic and linear functions, respectively. Make a rough sketch of these two functions on the same axes in the first quadrant.

Step 2: Find the points of intersection of the curves

To find the points of intersection, set the two functions equal to each other and solve for x: \(4x - x^2 = 4x - 4\) Simplify the equation and find the roots: \(-x^2 + 4 = 0\) Add \(x^2\) to both sides: \(x^2 = 4\) Taking square root of both sides gives us the points of intersection as x = -2 and x = 2. Since we are in the first quadrant, the point of intersection we are interested in is x = 2, so the curves intersect at point (2, 0).

Step 3: Set up the integral to find the area

The total area is the vertical distance between the two curves on the interval [0,2]. To find this distance in the first quadrant, integrate the difference between the two functions from 0 to 2: \(A = \int_0^2 (4x - x^2 - (4x - 4))dx\) Simplify the integral expression: \(A = \int_0^2 (-x^2 + 4)dx\)

Step 4: Solve the integral for the area

To find the area A, evaluate the definite integral: \(A = \left[ -\frac{1}{3}x^3 + 4x\right]_0^2\) Now, plug in the values of x = 0 and x = 2 and find the difference: \(A = \left(-\frac{1}{3}(2)^3 + 4(2) \right) - \left(-\frac{1}{3}(0)^3 + 4(0) \right)\) Simplify and calculate the area: \(A = \left(-\frac{8}{3} + 8 \right) - 0\) \(A = \frac{16}{3}\) So, the total area of the region in the first quadrant bounded by the given curves on the interval [0,2] is \(\frac{16}{3}\).

Key Concepts

Integrals

Integrals are a fundamental concept in calculus, related to the accumulation of quantities—essentially, they help us calculate areas under curves. When dealing with the area between two curves, particularly within defined limits as seen here, integrals become particularly useful. In simple terms, an integral allows us to sum up infinitely small quantities, which is necessary when finding the area within irregular shapes.
To find the area between two curves, we set up an integral that represents the difference between the top function and the bottom function over a specified interval. This is what we call the definite integral since it includes limits of integration that define the start and end points of our area of interest. In this case, those limits are from 0 to 2.
The symbolic representation of an integral looks like this: \[ \int_{a}^{b} f(x) \, dx \] Where \(f(x)\) is the function we're integrating over the interval from \(a\) to \(b\). The solution to the integral gives us the net area between the curve and the x-axis.
So, to compute the area between the two functions \(y = 4x - x^2\) and \(y = 4x - 4\) over the interval [0, 2], we compute:\[ A = \int_{0}^{2} (4x - x^2 - (4x - 4)) \, dx \] Resulting in calculating the area that lies above the curve \(y = 4x - 4\) and below \(y = 4x - x^2\), resulting in a net positive area.

Area under a curve

The concept of the "area under a curve" is integral to calculus and refers to the region contained between the graph of a function and the x-axis. In essence, this represents the aggregate of an infinite number of infinitesimally thin rectangles under the curve.
This concept becomes practical in real-life applications, such as finding distances, and volumes, or even consumer surplus in economics. In this particular exercise, finding the area under both curves means determining how much space lies between these two graphs within a specified interval, which is [0, 2].
To compute this, we essentially find the integral of the top curve and subtract the integral of the bottom curve:

• Top curve: \(y = 4x - x^2\)
• Bottom curve: \(y = 4x - 4\)

By doing this, we calculate only the "trapped" area between the two curves. The ultimate goal is to express the net result of these two integrations, yielding the complete area of interest, which comes out as \(\frac{16}{3}\). This value represents the finite portion of space that exists between these two functions on the interval from x = 0 to x = 2.

Quadratic functions

Quadratic functions, such as \(y = 4x - x^2\) in our example, are polynomial equations of degree two, which means their graph forms the shape of a parabola. More generally, the standard form of a quadratic function is:\[ y = ax^2 + bx + c \]For quadratic functions, the graph displays a characteristic symmetrical U or inverted U shape called a parabola.
The direction of the parabola (i.e., whether it opens upwards or downwards) depends on the coefficient \(a\). If \(a\) is positive, the parabola opens upwards; if \(a\) is negative, it opens downward. Here, \(a = -1\) within \(y = 4x - x^2\); hence, the parabola opens downward.
Key features of quadratic functions also include:

• The vertex, which in downward-facing parabolas is the maximum point, and can be found at \(x = -\frac{b}{2a}\) using the vertex formula.
• Intercepts, where the function crosses the x-axis (roots) and y-axis.

Quadratic functions are not only elegant mathematically but are also incredibly useful for modeling a wide variety of natural phenomena, from projectile motion in physics to business profit calculations. Understanding their properties allows us to solve problems involving area, such as intersecting another function, like in this exercise, seamlessly and accurately analyzing regions within a coordinate plane.