Problem 22 - Algebra 2

Question

Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ y=x^{5}-x^{2}+2.5 $$

Step-by-Step Solution

Answer

The area under the curve \(y = x^{5} - x^{2} + 2.5\) over the domain \(0 \leq x \leq 1\) is approximately 2.33 square units.

Step 1: Identify the function and the domain

The function to be integrated is \(y = x^{5} - x^{2} + 2.5\) over the domain \(0 \leq x \leq 1\). The target is to find the definite integral of this function over the specified domain.

Step 2: Setting up the integration

The problem translates to finding the definite integral of \(x^{5} - x^{2} + 2.5\) from 0 to 1, expressed as \(\int_{0}^{1} (x^{5} - x^{2} + 2.5) dx\).

Step 3: Evaluate the Integral

Using power rule for integration, \(\int_{0}^{1} (x^{5} - x^{2} + 2.5) dx = [1/6 * x^6 - 1/3 * x^3 + 2.5 * x]_{0}^{1}\). Substituting the limits of integration, result = \((1/6 - 1/3 + 2.5) - (0 - 0 + 0) = 2.33\)

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental method for finding antiderivatives of polynomial functions. It is the reverse process of differentiation using the power rule. In simpler terms, it helps us determine what function was differentiated to yield a given polynomial.
To apply the power rule, you increase the exponent by one and then divide by the new exponent.

• For example, for a term like \(x^n\), the integral is \(\frac{x^{n+1}}{n+1}\).
• Remember, this rule applies to all terms where the exponent \(n\) is not \(-1\).

In the exercise, we used the power rule to integrate each term of the polynomial \(x^5 - x^2 + 2.5\).

• The term \(x^5\) becomes \(\frac{x^6}{6}\).
• The term \(-x^2\) becomes \(-\frac{x^3}{3}\).
• The constant \(2.5\) integrates to \(2.5x\).

These integrated terms were then evaluated over the definite bounds to find the area under the curve.

Area Under a Curve

The area under a curve in a graph provides valuable information. In many practical contexts, it represents quantities like total distance, income, or population growth. For mathematical functions, it is determined using definite integrals, which compute the exact area between the curve and the x-axis over a specified interval.
A definite integral like \(\int_{a}^{b} f(x) \, dx\) computes this area precisely from \(x = a\) to \(x = b\).

• The result is sometimes a positive number, indicating an area above the x-axis.
• If the curve lies below the x-axis, the integral gives a negative value, which can be interpreted as negative area.

By calculating the definite integral of the function \(x^5 - x^2 + 2.5\) from 0 to 1 in our example, we found a positive area of 2.33.
This value represents the space enclosed by the curve, x-axis, and the vertical lines at \(x = 0\) and \(x = 1\).

Polynomial Functions

Polynomial functions are expressions consisting of variables raised to non-negative integer exponents and their coefficients. They form the basis for many mathematical models and analyses. These functions are not only smooth and continuous but also easy to work with due to well-established operational rules.

• They can be simple, like a linear function \(f(x) = mx + b\), or more complex with higher degrees.
• For instance, in our exercise, the polynomial \(y = x^5 - x^2 + 2.5\) is of degree 5.

Understanding polynomials allows one to predict and manipulate the behavior of more complicated functions. They are used across various scientific fields, from physics to economics, providing a method to describe real-world problems.
By breaking down each term of a polynomial separately, we can handle complex equations while applying rules like the power rule for integration effectively. This process simplifies solving integrals and finding areas under such curves.