Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=5 x-x^{2}, y=0\), between \(x=1\) and \(x=3\)

Short answer

The area of the region is approximately 11.333 square units.

Step-by-step solution

Identify the Equations and Bounds

The problem gives us two equations: 1. The function: \(y = 5x - x^2\) 2. The line: \(y = 0\)We are asked to find the area between these curves from \(x = 1\) to \(x = 3\). The line \(y = 0\) represents the x-axis.

Sketch the Region

To sketch the region, plot the function \(y = 5x - x^2\), which is a downward-opening parabola. Also, plot the line \(y = 0\) (x-axis). The region bounded by these is the area between the curve and x-axis from \(x = 1\) to \(x = 3\).

Visualize a Typical Slice

Imagine a vertical slice (a thin rectangle) from the curve \(y = 5x - x^2\) to the x-axis, somewhere between \(x = 1\) and \(x = 3\). The height of this slice is the value of the function at a given point \(x\), which is \(5x - x^2\).

Set Up the Integral

To find the area, integrate the height of the slices from \(x = 1\) to \(x = 3\). The integral to compute the area is:\[\int_{1}^{3} (5x - x^2) \, dx\]

Calculate the Integral

First, find the antiderivative of \(5x - x^2\):- The antiderivative of \(5x\) is \(\frac{5}{2}x^2\)- The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\)Plugging these into the integral, we get:\[\left[ \frac{5}{2}x^2 - \frac{1}{3}x^3 \right]_{1}^{3}\]Evaluate this from \(x = 1\) to \(x = 3\).

Evaluate the Definite Integral

Evaluate the antiderivative at the bounds:- At \(x = 3\): \(\frac{5}{2}(3)^2 - \frac{1}{3}(3)^3 = \frac{5}{2} \times 9 - \frac{1}{3} \times 27 = 22.5 - 9 = 13.5\)- At \(x = 1\): \(\frac{5}{2}(1)^2 - \frac{1}{3}(1)^3 = \frac{5}{2} \times 1 - \frac{1}{3} \times 1 = 2.5 - 0.333 = 2.167\)Thus, the area is \(13.5 - 2.167 = 11.333\).

Estimate the Area

To confirm our solution, we can estimate the area using geometric approximation or midpoint summation. Since the calculation uses definite integrals, an approximate method would yield a shoebox-like estimate around 11.5, close to our precise calculation of 11.333.

Key concepts

Area under a curve

The area under a curve in the context of calculus refers to the region between the graph of a function and the x-axis over a specified interval. This concept is essential when we want to determine the total "accumulated" value of a function over a range. In our exercise, the area we're interested in is bounded by the parabola defined by the equation \(y = 5x - x^2\) and the x-axis (where \(y = 0\)) from \(x = 1\) to \(x = 3\).

To visualize this, imagine the curve as a shape in the plane. We are interested in the specific section of this shape that "sits" between certain points along the x-axis. A vertical line at any point within these bounds forms the end of a "slice" of this shape. Integrating across these slices gives us the total area under the curve over the interval. This operation is central to understanding how we calculate such areas mathematically.

Whenever you need to determine such an area, you typically use the definite integral, a tool developed in calculus for this exact purpose. It allows you to sum up infinitesimally small segments of the curve to find the region beneath it.

Antiderivative

An antiderivative, often referred to as the indefinite integral, is a function whose derivative is the original function from which it is derived. In simpler terms, if you were to differentiate the antiderivative, you would get back the original function. It plays a crucial role in solving problems involving finding areas under curves.

In our step-by-step solution, we find the antiderivative of the function \(5x - x^2\). The antiderivative of \(5x\) is \(\frac{5}{2}x^2\), and the antiderivative of \(x^2\) is \(\frac{1}{3}x^3\). These calculations are performed by reversing the differentiation process.

The general method for finding an antiderivative involves using basic rules of integration, such as raising the power of \(x\) by one and dividing by the new power. The antiderivative is fundamental in computing the area under a curve, as it provides the necessary function to evaluate the definite integral.

Parabola

A parabola is a symmetric, U-shaped curve found in algebra and calculus, described by a quadratic equation. The general standard form of a parabola is \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Its shape is determined by the coefficient \(a\). When \(a\) is negative, the parabola opens downward, as is the case in our exercise with \(y = 5x - x^2\).

The equation \(y = 5x - x^2\) thus describes a downward-opening parabola with a vertex that can be found by using the vertex formula \(x = -\frac{b}{2a}\). Understanding the shape and orientation of the parabola helps in visualizing the area we wish to calculate.

In calculus problems, especially involving integration, recognizing the type of curve you are working with (like a parabola) can guide you in setting up correct integrals and anticipating the kind of solution you might expect.

Definite Integral Evaluation

Evaluating a definite integral is the process of calculating the actual number that represents the area under a curve within specified bounds. The definite integral of a function \(f(x)\) from \(a\) to \(b\) is denoted \(\int_{a}^{b} f(x) \, dx\). This operation results in a real number that represents the total accumulated area or quantity described by the function.

For our parabolic function \(y = 5x - x^2\), the integral to evaluate the area under the curve from \(x = 1\) to \(x = 3\) is \(\int_{1}^{3} (5x - x^2) \, dx\). This integral provides the mechanism to sum up the "slices" of area discussed earlier and output a precise value, 11.333 in this case.

To perform the evaluation:

• Find the antiderivative of the function: \(\frac{5}{2}x^2 - \frac{1}{3}x^3\).
• Determine the value of this antiderivative at the bounds, i.e., substitute \(x = 3\) and \(x = 1\) into the antiderivative.
• Subtract the lower bound result from the upper bound result: \(13.5 - 2.167 = 11.333\).

This final calculation provides the exact area under the curve for the given interval, showcasing the power of calculus techniques in solving complex real-world problems.