Problem 25 - Calculus: Early Transcendental Functions

Question

Sketch the area corresponding to the integral. $$\int_{1}^{2}\left(x^{2}-x\right) d x$$

Step-by-Step Solution

Answer

To sketch the area corresponding to the integral \(\int_{1}^{2}(x^{2}-x) dx\), plot the function, calculate the definite integral, then shade the region under the curve and above the x-axis between x = 1 and x = 2. This shaded area represents the value of the integral.

Step 1: Understand the Function

Given is the function \(x^{2} - x\). This is a quadratic function which opens upwards. Identifying the type of the function is critical, as different functions will have different shaped curves when graphed.

Step 2: Plot the Function

Plot the function \(x^{2} - x\) on an appropriate range. The x-values are given as 1 to 2 in the integral. Hence, use this as the range to plot the function.

Step 3: Calculate the Integral

Calculate the integral of \(x^{2} - x\) from 1 to 2. In order to calculate a definite integral, find the antiderivative of the function first, then substitute the upper and lower limits of the integral into this antiderivative function. The formula for the definite integral is: \(\int_{a}^{b} f(x) dx = F(b) - F(a)\), where F(x) is the antiderivative of f(x). These calculations result in a numerical value, representing the area under the curve from 1 to 2.

Step 4: Sketch the Area under the Curve

After the function is plotted and integrated, sketch the area under the curve which corresponds to the interval from 1 to 2. In this case, shade the region under the curve and above the x-axis between x = 1 and x = 2 to represent the integral.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial that is defined by the form \( ax^2 + bx + c \). It is characterized by its U-shaped graph, known as a parabola. In this exercise, the specific quadratic function provided is \( x^2 - x \). Here, the coefficient \(a\) is positive, meaning the parabola opens upwards.
Understanding the structure of a quadratic function is essential as it helps in visualizing the curve, which affects how you interpret the integral. Important features of quadratic functions include:

• Vertex: The highest or lowest point of the parabola. For the function \( x^2 - x \), you can use \( -\frac{b}{2a} \) to find the x-value of the vertex.
• Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetric halves. For this function, the axis of symmetry is \( x = 0.5 \).
• Roots: Points where the curve intersects the x-axis. Solve \( x^2 - x = 0 \) to find \( x = 0 \) and \( x = 1 \).

The behavior of the quadratic function between these roots influences the calculation of the area under its curve, as it defines the integral's boundaries.

Antiderivative

The antiderivative, also known as the indefinite integral, is a function whose derivative gives back the original function. In the context of definite integrals, finding the antiderivative is a crucial step.
For the given function \( x^2 - x \), the antiderivative is computed as follows:
\[ \int (x^2 - x) \, dx = \frac{x^3}{3} - \frac{x^2}{2} + C \]
Where \( C \) represents the constant of integration. However, when dealing with definite integrals, \( C \) cancels out as you evaluate the difference \( F(b) - F(a) \).
The general steps to find an antiderivative include:

• Identify each term in the function.
• Apply the reverse of differentiation rules. For polynomials, increase the power by one and divide by the new power.

Understanding how to find the antiderivative helps in evaluating the area under a curve between specified limits, which is what a definite integral represents.

Area Under Curve

The concept of finding the area under a curve involves evaluating a definite integral over a specific interval. The area represents the accumulation of values of the function, often interpreted in physical terms like distance or total quantity.
To calculate the area under the curve \( x^2 - x \) from 1 to 2:

• First, find the antiderivative: \( F(x) = \frac{x^3}{3} - \frac{x^2}{2} \).
• Evaluate the antiderivative at the upper limit, \( F(2) \), and the lower limit, \( F(1) \).
• Subtract these results: \( F(2) - F(1) \).

This process gives the numerical value of the area under the curve between \( x = 1 \) and \( x = 2 \), which effectively measures the space beneath the quadratic curve and above the x-axis.
This is a key concept in calculus, as the area under the curve can be used to solve a variety of practical problems.