Chapter 7: Problem 113
Find a substitution \(w\) and a constant \(k\) so that the integral has the form \(\int k e^{i t} d w\). $$\int \frac{z^{2} d z}{e^{-z^{3}}}$$
Short Answer
Expert verified
Let \(w = z^3\) and \(k = \frac{1}{3}\). The integral becomes \(\int \frac{1}{3} e^{w} dw\).
Step by step solution
01
Identify the Structure of the Integral
We want to transform the integral \(\int \frac{z^{2} dz}{e^{-z^{3}}} \) into the form \(\int k e^{i t} dw\). Notice that the denominator can be re-written as a function of the form \(e^u\). This suggests that \(w\) should be related to \(z^3\) to modify the expression in a way that simplifies the exponential term.
02
Choose the Substitution for the Integral
Let \( w = z^3 \). Then, the differential \( dw = 3z^2 dz \). This choice simplifies the expression in the denominator because it allows us to express \(e^{-z^3}\) as \(e^{-w}\).
03
Adjust the Integral Using the Substitution
Using the substitution from Step 2, express \(dz\) in terms of \(dw\): we have that \(dz = \frac{dw}{3z^2}\). Thus, \(z^2 dz = \frac{1}{3} dw\). SUBSTITUTE this into the integral:\[ \int \frac{z^2 dz}{e^{-z^3}} = \int \frac{1}{3} \frac{dw}{e^{-w}} \].
04
Simplify the Integral Expression
Recognizing \( e^{-z^3} \) is \( e^{-w} \), rewrite the integral:\[ \int \frac{1}{3} e^{w} dw \].This can be transformed further to separate constant terms, making it:\[ \frac{1}{3} \int e^{w} dw \].
05
Match the Integral to the Desired Form
We need the integral to match \(\int k e^{i t} dw\). Comparing this with \(\frac{1}{3}\int e^{w} dw\), we can let \(k = \frac{1}{3}\) and understand that the \(i t\) component aligns with \( w \) as a simpler real exponential. Thus, the integral is already simplified and in a closely desired form.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions represent a unique form of mathematical functions where a constant base is raised to a variable exponent, such as in the case of functions like \( e^x \). These functions frequently occur in various branches of mathematics, especially calculus, due to their fundamental growth properties.
Exponential functions have several key characteristics:
Exponential functions have several key characteristics:
- They are always positive if the base is greater than zero.
- They have a constant rate of growth or decay proportional to their current value.
- They are continuous and differentiable for all real numbers.
Definite and Indefinite Integrals
When solving calculus problems, you'll often encounter two types of integrals: definite and indefinite. Integrals are essential for calculating areas under curves, among other applications.
- Indefinite Integrals: These represent the antiderivative of a function and are crucial in operations involving the recovery of original functions from their derivatives. They are typically expressed with the inclusion of a constant of integration, denoted often by \( C \).
- Definite Integrals: These are concerned with the actual computation of area under a curve between two specific points, providing a numerical value that reflects this area.
Change of Variable Technique
The change of variable technique, commonly known as integration by substitution, is a powerful method used to simplify integral calculations. This approach involves changing the variable of integration to make an integral easier to evaluate.
- The method usually begins by making a suitable substitution of variables, such as \( w = z^3 \) from the exercise, which simplifies the integrals by transforming complex expressions into simpler ones.
- The substitution affects both the variable and the differential, hence always ensuring the correct transformation of \( dz \) to \( dw \), as seen with the expression \( dw = 3z^2 dz \).
- The goal of this technique is to rewrite the integral in terms of the new variable, often turning a challenging integral into a more straightforward one, essentially simplifying the required computations.