Problem 50 Evaluate the integrals in Exerci... [FREE SOLUTION]

Chapter 5: Problem 50

Evaluate the integrals in Exercises $17-50$ $$ \int \frac{x}{(2 x-1)^{2 / 3}} d x $$

Short Answer

Expert verified

The integral of $ \int \frac{x}{(2x-1)^{2/3}} dx $ is $ \frac{3}{16}(2x-1)^{4/3} + \frac{3}{4}(2x-1)^{1/3} + C $.

Step by step solution

Choose a suitable substitution

We start by choosing a substitution that will simplify the integral. Let $ u = 2x - 1 $. This implies that $ du = 2 dx $ or equivalently, $ dx = \frac{du}{2} $. This substitution simplifies both the expression under the integral and its differential.

Rewrite the integral in terms of $ u $

By substituting $ 2x = u + 1 $, we have $ x = \frac{u + 1}{2} $. Replace $ x $ and $ dx $ in the integral: \[ \int \frac{\frac{u+1}{2}}{u^{2/3}} \cdot \frac{du}{2} = \int \frac{u+1}{2} \cdot \frac{1}{u^{2/3}} \cdot \frac{du}{2} \].

Simplify the integrand

Simplify the expression: \[ \frac{1}{4} \int \left( \frac{u+1}{u^{2/3}} \right) du = \frac{1}{4} \left( \int \frac{u}{u^{2/3}} du + \int \frac{1}{u^{2/3}} du \right) \]. This further simplifies to: \[ \frac{1}{4} \int u^{1 - 2/3} du + \frac{1}{4} \int u^{-2/3} du \]. Both terms are now ready for integration.

Integrate each term separately

Integrate the first term: \[ \int u^{1/3} du = \frac{3}{4} u^{4/3} + C_1 \].Integrate the second term: \[ \int u^{-2/3} du = 3u^{1/3} + C_2 \].Combine the results of these integrals.

Evaluate the definite integral

Combine the integrated terms: \[ \frac{1}{4} \left( \frac{3}{4}u^{4/3} + 3u^{1/3} \right) + C \].This simplifies to:\[ \frac{3}{16} u^{4/3} + \frac{3}{4} u^{1/3} + C \]. Replace $ u $ back with $ 2x-1 $ to get the final answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

definite integral

In calculus, a definite integral is a way to calculate the accumulation of quantities, such as area under a curve. This type of integral has limits, denoting the interval over which integration is performed. Here, boundaries are not explicitly given, but typically, evaluating a definite integral involves boundaries.

The definite integral of a function from point A to point B is represented as $ \int_{A}^{B} f(x) \, dx $.
It provides a finite number, representing the net area between the x-axis and the curve from these two points.
In contrast to the indefinite integral, a definite integral does not include a constant of integration, as the result is an exact value, not a family of functions.

Understanding definite integrals is crucial in many fields. They are often used in physical sciences to calculate quantities like work and energy. The methodology described later could aid immensely in finding the solution to definite integrals when limits are defined.

substitution method

The substitution method is a powerful technique in integral calculus used to simplify and solve integrals. It works on the principle of substituting a part of the original integral with a new variable to make the equation easier to evaluate.

Start by choosing a portion of the integrand to replace with a new variable $ u $.
Express $ dx $ in terms of $ du $ using differentiation.
Rewrite the original integral in terms of $ u $ and simplify.
Integrate with respect to $ u $, then substitute back to the original variable.

This method effectively transforms a complex integral into a simpler one, making the calculation more manageable. It is particularly useful for integrals involving composite functions, enabling solutions that would be otherwise difficult to achieve through direct methods.Learning to pick the right substitution is a skill that comes with practice and is key to utilizing this method effectively.

integral calculus

Integral calculus is a branch of mathematics focusing on the concept of integration, which is essentially the inverse of differentiation. It involves finding the antiderivative of a function.

Indefinite integrals, written as $ \int f(x) \, dx $, yield a general form without specific bounds. This includes the constant of integration $ C $.
It is used to solve problems involving accumulation, such as finding areas, volumes, and sums of infinite series.
Understanding its rules and techniques is crucial for progressing through calculus and applying it to various problems.

Integral calculus opens a world of solving real-world problems by providing tools to determine accumulated values over intervals. Like the substitution method described, various techniques improve problem-solving efficiency, enhancing the depth of analysis in mathematical and physical contexts.Mastery of integral calculus is central to advanced studies in mathematics and sciences, forming the backbone of more complex problem-solving endeavors.

91影视

Evaluate the integrals in Exercises \(17-50\) $$ \int \frac{x}{(2 x-1)^{2 / 3}} d x $$

Short Answer

Step by step solution

Choose a suitable substitution

Rewrite the integral in terms of \( u \)

Simplify the integrand

Integrate each term separately

Evaluate the definite integral

Key Concepts

definite integral

substitution method

integral calculus

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