/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Evaluate the indefinite integral... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the indefinite integrals (use the substitutions in parentheses, when given): \(\int(3 x+1)^{5} d x \quad(u=3 x+1)\)

Short Answer

Expert verified
\(\frac{(3x+1)^6}{18} + C\)

Step by step solution

01

Identify the substitution

We start by using the given substitution. Let \( u = 3x + 1 \). This will change our integral in terms of \( u \).
02

Differentiate the substitution

Differentiate the equation \( u = 3x + 1 \) with respect to \( x \) to find \( du/dx \). This gives us \( \frac{du}{dx} = 3 \), or equivalently, \( du = 3 \, dx \).
03

Solve for dx

From \( du = 3 \, dx \), we solve for \( dx \) in terms of \( du \). So, \( dx = \frac{du}{3} \).
04

Substitute in the integral

Replace \((3x+1)^5\) with \(u^5\) using \( u = 3x + 1 \), and replace \( dx \) with \( \frac{du}{3} \). The integral becomes: \( \int u^5 \frac{du}{3} \).
05

Simplify the integral

Factor out the constant \( \frac{1}{3} \) from the integral. This gives us \( \frac{1}{3} \int u^5 \, du \).
06

Integrate

Perform the integration using the power rule for integrals. The integral of \( u^5 \) is \( \frac{u^6}{6} + C \), where \( C \) is the integration constant. So, the result is \( \frac{1}{3} \cdot \frac{u^6}{6} + C = \frac{u^6}{18} + C \).
07

Substitute back the original variable

Substitute back \( u = 3x + 1 \) into the solution: \( \frac{(3x+1)^6}{18} + C \). This is the antiderivative of the original integral.

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Key Concepts

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

Integration by Substitution
Integration by substitution is a technique used to evaluate integrals more easily by simplifying the integrand. When dealing with complex functions, substitution helps make the integration easier by replacing a complicated part of the integrand with a single variable. This process is somewhat similar to using the chain rule in differentiation but in reverse.

Here's a quick summary of how to apply it:
  • Identify a part of the integrand that's complicated and requires simplification. Choose the substitution such that the derivative of this part is also present in the integral.
  • Re-express the integral in terms of the new variable.
  • Don't forget to change the differential (e.g., change from \( dx \) to \( du \)).
  • Perform the integration in the simplified form and then substitute back the original variable to obtain the final result.
In the example in the exercise, we used \( u = 3x + 1 \) to simplify \((3x+1)^5\), making the process of integration more straightforward. This is a key tool in calculus that eases the complexity of integration for more challenging functions.
Antiderivatives
Antiderivatives are essentially the reverse of derivatives. When you find the antiderivative of a function, you are finding a new function whose derivative is the given function. Antiderivatives are also known as "indefinite integrals." This is because they produce a family of functions defined by the integration constant \( C \). The general form is:
  • If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
Returning to the exercise, when integrating \( u^5 \), the antiderivative is \( \frac{u^6}{6} + C \). This means the derivative of this function (with respect to \( u \)) would return the original function \( u^5 \). Always include the constant \( C \) because indefinite integrals yield many possible solutions differing by a constant value.

This is fundamental because when solving real-world problems, these constants can represent initial conditions or starting values of a system being modeled.
Calculus
Calculus is the branch of mathematics that studies continuous change. It has two main branches: differential calculus and integral calculus. Integral calculus, which is our focus here, deals with finding the size of something, like the area under a curve, by summing infinitely small things to find a whole.

Key concepts in calculus that were used in solving the problem include:
  • Indefinite Integrals: Used to find the antiderivative of functions.
  • Integration by Substitution: A method to simplify complex integrals.
  • Power Rule: An important rule used in integration to determine the antiderivative of powers of \( x \).
Understanding calculus is essential for grasping how different mathematical quantities change and interact. It's widely applicable in physics, engineering, economics, and beyond, providing solutions to real-world problems through mathematical modeling and analysis.

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