Question

Evaluate the definite integral. $$ \int_{-1}^{1}(2 t-1)^{2} d t $$

Short answer

The value of the definite integral is \( \frac{8}{3} \).

Step-by-step solution

Identify the function to be integrated

The function to be integrated is \( (2t-1)^2 \). This function is a square of a binomial, and it can be expanded to simplify the integration process.

Expand the function

Expanding the square of a binomial, the function \( (2t-1)^2 \) becomes \( 4t^2 - 4t + 1 \). From this step, it can be seen that the integration process will involve finding the antiderivative of a polynomial function.

Integrate the polynomial

The antiderivative of the polynomial function \( 4t^2 - 4t + 1 \) is \( \frac{4}{3}t^3 - 2t^2 + t \). This is deduced from the power rule of integration, which states the antiderivative of \( t^n \) is \( \frac{1}{n+1}t^{n+1} \).

Evaluate the Definite Integral

The definite integral is the difference in the antiderivative evaluated at the given limits of integration. So first, plug \(t = 1\) into \( \frac{4}{3}t^3 - 2t^2 + t \) to get \( \frac{4}{3} - 2 + 1 = \frac{1}{3} \). Then plug \(t = -1\) in and get \( -\frac{4}{3} - 2 -1 = -\frac{7}{3} \). The definite integral is then \( \frac{1}{3} - (-\frac{7}{3}) = \frac{8}{3} \).

Key concepts

Polynomial Function

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In simpler terms, a polynomial is a combination of terms like constants, variables raised to various powers, and these powers can be positive integers or zero.
For example, the function \(4t^2 - 4t + 1\) is a polynomial of degree 2 because the highest power of \(t\) is 2. This concept is foundational in calculus and algebra since many mathematical operations, including integration, often deal with polynomials.

• They can be expanded or simplified using algebraic operations.
• Each term in a polynomial is a product of a constant and a variable raised to a whole number power.
• Polynomial functions are smooth and continuous, meaning there's no abrupt change in their graphs.

Understanding the structure of polynomials makes it easier to perform operations like finding derivatives and integrals.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. It effectively "undoes" differentiation. When a function is differentiated and then integrated, it yields the original function plus a constant because integration is the reverse process of differentiation.
The antiderivative of a polynomial involves reversing the differentiation process by adding one to the exponent and adjusting the coefficient accordingly. For instance, when you have a polynomial \(4t^2 - 4t + 1\), its antiderivative is \(\frac{4}{3}t^3 - 2t^2 + t + C\), where \(C\) represents any constant since integration results add an indefinite constant.

• Antiderivatives are essential in finding areas under curves, which is the core idea of definite integration.
• They provide a way to return to the original function from its rate of change.
• Every continuous function has an antiderivative.

Power Rule of Integration

The power rule of integration is a straightforward method for finding the integral of polynomial expressions. This rule is applicable to terms of the form \(t^n\), where \(n\) is any real number except \(-1\).
According to the power rule, the antiderivative of \(t^n\) is \(\frac{1}{n+1}t^{n+1}\) plus a constant \(C\). This rule provides a quick and efficient way to solve integration problems involving polynomials.
For example, considering the polynomial \(4t^2\), applying the power rule gives you \(\frac{4}{3}t^3\), because you add 1 to the exponent 2 (making it 3) and divide the coefficient 4 by this new exponent 3.

• The power rule simplifies the process of integrating polynomial functions.
• Each integral result may have an added constant, representing infinite possible original functions.
• The power rule is a fundamental tool used in calculus for solving integrals quickly and accurately.

Understanding and applying this rule helps with efficiently solving definite integrals, as seen in the original exercise.