Problem 13 - Calculus Early Transcendentals

Question

Evaluate the following integrals. Include absolute values only when needed. \(\int_{0}^{3} \frac{2 x-1}{x+1} d x\)

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int_{0}^{3} \frac{2x-1}{x+1} dx\). Answer: The integral \(\int_{0}^{3} \frac{2x-1}{x+1} dx\) is equal to \(0\).

Step 1: Finding the antiderivative

To find the antiderivative of the given function, we can first try to simplify it. We notice that the denominator can be factored from the numerator, which can help in simplification and integration. \(\frac{2x-1}{x+1} = \frac{2(x+1)-2-x}{x+1} = 2 - \frac{2+x}{x+1} = 2 - \frac{2(1+x)}{2(x+1)}\) Now we can integrate the simplified function term by term: \(\int \frac{2x-1}{x+1} dx = \int (2 - \frac{2(1+x)}{2(x+1)}) dx = 2\int dx - \int \frac{2(1+x)}{2(x+1)} dx\) Notice that the second term can be integrated using the substitution method. Let \(u = x+1 \Rightarrow du = dx\). Thus, the integral becomes: \(= 2\int dx - \int \frac{2(u)}{2(u)} du\) Now we can integrate each term: \(= 2x - 2\int du\) \(= 2x - 2u + C\) Now, substitute \(u\) back to terms of \(x\): \(= 2x - 2(x+1) + C = 2x - 2x - 2 + C = -2 + C\) The antiderivative of \(\frac{2x-1}{x+1}\) is \(F(x) = -2 +C\).

Step 2: Evaluating the integral using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that: \(\int_{a}^{b} f(x) dx = F(b) - F(a)\) We have found the antiderivative \(F(x) = -2+C\). Now, we need to evaluate the integral over the given interval \([0, 3]\): \(\int_{0}^{3} \frac{2x-1}{x+1} dx = F(3) - F(0)\) Since we have a constant C for both F(3) and F(0), the constant cancels out, resulting in: \(= (-2) - (-2) = -2 + 2 = 0\) So, the integral \(\int_{0}^{3} \frac{2x-1}{x+1} dx = 0\).

Key Concepts

Antiderivative

Antiderivatives are essential in the process of integration. When you integrate a function, you are essentially finding its antiderivative. An antiderivative is a function that, when differentiated, gives back the original function. For example, if you have a function \( f(x) = 2x \), its antiderivative could be \( F(x) = x^2 + C \), where \( C \) represents any constant, reflecting the indefinite nature of the integration.In definite integrals, which have specified limits, this constant \( C \) becomes irrelevant, because it cancels out during evaluation (as seen in the original exercise). When you find the antiderivative for an expression and evaluate it over a given interval, you can calculate the definite integral without uncertainty, achieving precisely what was obtained in the solution.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the disciplines of differentiation and integration, two central operations in calculus. It provides a way to calculate definite integrals \(\int_{a}^{b} f(x) \, dx \) without directly summing the areas but instead identifying antiderivatives.This theorem states that to evaluate the definite integral of a function between two points \( a \) and \( b \), you can find the antiderivative function \( F(x) \):

• Compute \( F(b) \): plug the upper limit into the antiderivative.
• Compute \( F(a) \): plug the lower limit into the antiderivative.
• Subtract \( F(a) \) from \( F(b) \).

In the example, we used the theorem to find that the integral of \( \frac{2x-1}{x+1} \) from \( 0 \) to \( 3 \) resulted in zero, highlighting the elegant cancellation that can sometimes occur with constants in antiderivatives.

Integration by Substitution

Integration by substitution is an essential technique used to simplify complicated integrals, similar to the reverse of the chain rule in differentiation. By transforming the integrals into a simpler form, you can more easily find antiderivatives.For instance, when encountering an integral like \( \int \frac{2(1+x)}{2(x+1)} dx \), substitution simplifies the process:- Letting \( u = x + 1 \) changes \( dx \) to \( du \), and often reduces the complexity of the function under the integral.- Substitute back the value after evaluating the simpler integral.This technique essentially changes the variable of integration to make the integral easier to evaluate. This was used effectively in the given problem, allowing us to find an antiderivative and evaluate the integral in a straightforward manner. The continuous practice of such techniques builds intuition and skill, which is invaluable in tackling more challenging calculus problems.