Chapter 5: Problem 6
Evaluate the definite integral two ways: first by a \(u-\) substitution in the definite integral and then by a \(u\) -substitution in the corresponding indefinite integral. $$ \int_{1}^{2}(4 x-2)^{3} d x $$
Short Answer
Expert verified
The value of the definite integral is 80.
Step by step solution
01
Define the Substitution
Let us set \( u = 4x - 2 \). Then, differentiate to get \( \frac{du}{dx} = 4 \), or \( du = 4 \, dx \). Since \( dx = \frac{1}{4} \, du \), our integral becomes \( \int (4x - 2)^3 dx = \int u^3 \cdot \frac{1}{4} du \).
02
Adjust Limits of Integration
Change the limits of integration according to the substitution. When \( x = 1 \), \( u = 4(1) - 2 = 2 \). When \( x = 2 \), \( u = 4(2) - 2 = 6 \). Now, the definite integral is \( \int_{2}^{6} \frac{1}{4} u^3 \, du \).
03
Evaluate the Definite Integral
Evaluate the integral \( \int \frac{1}{4} u^3 du \). This becomes \( \frac{1}{4} \left[ \frac{u^4}{4} \right] \). Evaluate this from \( u = 2 \) to \( u = 6 \). It equals \( \frac{1}{16} [(6)^4 - (2)^4] = \frac{1}{16} [1296 - 16] = \frac{1}{16} \times 1280 = 80 \).
04
Find Indefinite Integral with Substitution
For the indefinite integral \( \int (4x - 2)^3 dx \), first substitute \( u = 4x - 2 \) and \( dx = \frac{1}{4} du \), leading to \( \int \frac{1}{4} u^3 du = \frac{1}{16} \cdot \frac{u^4}{4} + C = \frac{u^4}{16} + C \).
05
Convert Back and Determine Constant of Integration
After finding the indefinite integral \( \frac{u^4}{16} + C \), replace \( u = 4x - 2 \) back into the function, obtaining \( \frac{(4x - 2)^4}{16} + C \).
06
Evaluate Indefinite Integral from 1 to 2
Now substitute \( x = 1 \) and \( x = 2 \) to determine the definite integral. For \( x = 2 \), this is \( \frac{(4(2)-2)^4}{16} = \frac{6^4}{16} = 81 \), and for \( x = 1 \), this is \( \frac{(4(1)-2)^4}{16} = \frac{2^4}{16} = 1 \). Thus, the result is \( 81 - 1 = 80 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
U-Substitution
One powerful technique for solving integrals is "U-Substitution," also known simply as substitution. It's a method that is especially useful for simplifying integrals by changing variables. Here's how it works:- The goal of U-Substitution is to replace a part of the integral with a single variable, usually denoted as \( u \). This makes the integral easier to solve.- First, identify a portion of the integrand (the function being integrated) that can be substituted. You then let \( u \) equal this portion.- Differentiate \( u \) to find \( du \) in terms of \( dx \). In our example, \( u = 4x - 2 \), hence \( du = 4 \, dx \).- The original width of each infinitesimal slice of \( dx \) is now expressed as \( \frac{1}{4} \, du \). Then, replace every instance of the original variable with \( u \) in the integrand.This method is extremely helpful for tackling complex integrals, leading you to simplified expressions that are much easier to integrate.
Indefinite Integral
An indefinite integral represents a family of functions and is opposite to taking a derivative. This process provides the antiderivative or original function whose derivative gives the integrand.- When solving for an indefinite integral, you'll often see a constant \( C \) added after integration. This constant is necessary because the process of differentiation removes constants.- For example, if you have a function \( f(u) = u^3 \), its integral is \( \int u^3 \, du = \frac{u^4}{4} + C \).When using U-Substitution in the context of indefinite integrals, remember to replace the substitution variable back to the original variable at the end of integration.- In our example, once the indefinite integral \( \int (4x - 2)^3 \, dx \) is calculated using \( u \), the final expression should return to the variable \( x \), yielding \( \frac{(4x - 2)^4}{16} + C \).Understanding indefinite integrals helps in solving equations that require rebuilding original functions from their slopes or rates of change.
Calculus
Calculus is a fascinating branch of mathematics that deals with change and motion. It is divided into two main areas: differentiation and integration.- Differentiation focuses on finding rates of change. It's about understanding how a function changes at any given point, like how fast an object travels at a specific moment.- Integration is essentially the reverse process. While differentiation provides the rate, integration seeks the original function from this rate. It's akin to deducing the distance covered from a known speed.Within calculus, we regularly rely on concepts such as definite and indefinite integrals:- **Definite integrals** have limits and provide numerical values representing the area under a curve between two points.- **Indefinite integrals** are about finding general expressions for antiderivatives and include an arbitrary constant \( C \).Mastering calculus involves understanding these core concepts, along with techniques like U-Substitution, to effectively solve complex mathematical problems.