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Chapter 5: Problem 21

Use the Substitution Formula in Theorem 7 to evaluate the integrals. $$\int_{0}^{1}\left(4 y-y^{2}+4 y^{3}+1\right)^{-2 / 3}\left(12 y^{2}-2 y+4\right) d y$$

Short Answer

Expert verified
The integral evaluates to 3.

Step by step solution

01

Identify parts of the integral for substitution

To use substitution, identify the inner function of the composite function. Here, the expression inside the parenthesis is a good candidate for substitution: let \( u = 4y - y^2 + 4y^3 + 1 \).
02

Differentiate the substitution function

Find the derivative of \( u \) with respect to \( y \). Start by calculating \( \frac{du}{dy} \):\[ \frac{du}{dy} = 4 - 2y + 12y^2 \].
03

Adjust the integral

Rewrite the differential \( dy \) in terms of \( du \) using the derivative obtained:\( dy = \frac{1}{4 - 2y + 12y^2} du \). Substitute \( u \) and \( dy \) into the integral.
04

Substitute in the integral

Replace the original integral with terms involving \( u \):\[ \int_{0}^{1} u^{-2/3} \frac{12y^2 - 2y + 4}{4 - 2y + 12y^2} du = \int u^{-2/3} du. \]Here, the terms \(12y^2 - 2y + 4\) and \(4 - 2y + 12y^2\) cancel each other.
05

Solve the simplified integral

Now, integrate \( u^{-2/3} \):\[ \int u^{-2/3} du = \frac{u^{1/3}}{1/3} + C = 3u^{1/3} + C. \]
06

Back-substitute and evaluate limits

Replace \( u \) back with \( 4y - y^2 + 4y^3 + 1 \) in the integrated result, obtaining:\[ 3(4y - y^2 + 4y^3 + 1)^{1/3}. \]Evaluate this expression from 0 to 1:- At \(y = 0\): \( 3(4 \times 0 - 0^2 + 4 \times 0^3 + 1)^{1/3} = 3 \times 1^{1/3} = 3 \).- At \(y = 1\): \( 3(4 \times 1 - 1^2 + 4 \times 1^3 + 1)^{1/3} = 3 \times 8^{1/3} = 3 \times 2 = 6 \).
07

Calculate the definite integral

Subtract the evaluated limits to find the integral: \( 6 - 3 = 3 \).
08

Final Result

The definite integral evaluates to 3 after completing the substitution and simplification.

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool in integral calculus that simplifies the integration process, especially for composite functions. The core idea is to substitute a part of the integrand with a new variable, typically denoted as \( u \). This substitution should transform the integral into a simpler form. In practice:

  • Identify a part of the integral that resembles a composite function.
  • Let this part be \( u \), and then find its derivative \( \frac{du}{dy} \).
  • Use this derivative to express \( dy \) in terms of \( du \).
  • Rewrite the integral in terms of \( u \), effectively simplifying it.
  • After integrating, substitute back to the original variable.
By using substitution, the integration process is often made more straightforward, especially when dealing with functions that may seem complex at first glance.
Definite Integrals
Definite integrals are a fundamental concept in calculus used to calculate the area under a curve between two specific points. Unlike indefinite integrals, definite integrals have limits of integration, usually labeled as \( a \) and \( b \). The main steps in solving a definite integral are:
  • Evaluate the integral expression within the provided limits.
  • Substitute the limits into the integrated function to find the values at these points.
  • Subtract the value at the lower limit from the value at the upper limit to find the accumulated area.
Definite integrals are particularly useful in various applications, such as finding the total distance traveled over time or the total accumulation of a quantity. The final results from these calculations provide tangible values that depict how quantities change within a given interval.
Composite Functions
Composite functions are functions formed when one function is applied within another. In mathematics notation, if \( f(x) \) and \( g(x) \) are functions, the composite function \( f(g(x)) \) occurs by replacing the variable in \( f \) with \( g(x) \). This can create complex expressions but at the same time provide flexibility in representing real-world phenomena.

To deal with composite functions in integration, substitution is often necessary. This simplification becomes crucial when you need to handle nested functions effectively. By tackling inside-out, that is, focusing on the most embedded functions first, integration becomes more manageable.

Understanding composite functions allows one to bridge various mathematical expressions and apply calculus tools effectively, making it critical to numerous fields like physics, engineering, and economics.

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Most popular questions from this chapter

Find \(d y / d x\).$$y=\int_{\tan x}^{0} \frac{d t}{1+t^{2}}.$$

It would be nice if average values of integrable functions obeyed the following rules on an interval \([a, b].\) a. \(\operatorname{av}(f+g)=\operatorname{av}(f)+\operatorname{av}(g)\) b. \(\operatorname{av}(k f)=k \operatorname{av}(f) \quad\) (any number \(k\) ) c. \(\operatorname{av}(f) \leq \operatorname{av}(g) \quad\) if \(\quad f(x) \leq g(x) \quad\) on \(\quad[a, b]\) Do these rules ever hold? Give reasons for your answers.

Find the linearization of $$g(x)=3+\int_{1}^{x^{2}} \sec (t-1) d t$$at \(x=-1.\)

You will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$f(x)=x+\sin (2 x), \quad g(x)=x^{3}$$

Suppose that \(f\) has a positive derivative for all values of \(x\) and that \(f(1)=0 .\) Which of the following statements must be true of the function $$g(x)=\int_{0}^{x} f(t) d t ?$$. Give reasons for your answers. a. \(g\) is a differentiable function of \(x\) b. \(g\) is a continuous function of \(x\). c. The graph of \(g\) has a horizontal tangent at \(x=1\) d. \(g\) has a local maximum at \(x=1\) e. \(g\) has a local minimum at \(x=1\) f. The graph of \(g\) has an inflection point at \(x=1\) g. The graph of \(d g / d x\) crosses the \(x\) -axis at \(x=1\)
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