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Chapter 7: Problem 43

Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the function and its antiderivative (take \( C = 0 \)). \( \displaystyle \int xe^{-2x} dx \)

Short Answer

Expert verified
The antiderivative is \(-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\).

Step by step solution

01

Identify Integration Method

Notice that the integral \( \int xe^{-2x} \, dx \) involves a polynomial \( x \) and an exponential \( e^{-2x} \). This suggests using integration by parts, where we let \( u = x \) and \( dv = e^{-2x} \, dx \).
02

Determine \( du \) and \( v \)

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Calculate \( du \) as the derivative of \( u \), which is \( du = dx \), and find \( v \) by integrating \( dv \):1. \( du = dx \)2. Integrate \( dv = e^{-2x} \, dx \) - \( v = \int e^{-2x} \, dx = \frac{-1}{2} e^{-2x} \)
03

Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[\int xe^{-2x} \, dx = x\left(\frac{-1}{2}e^{-2x}\right) - \int \left(\frac{-1}{2}e^{-2x}\right) \, dx\]Simplify:\[= -\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x} \, dx\]
04

Evaluate Remaining Integral

Complete the integration for \( \int e^{-2x} \, dx \):\[= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\]where \( C \) is the constant of integration.
05

Confirm Result with Graphs

Graph both the original function \( xe^{-2x} \) and its antiderivative \( -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \) with \( C = 0 \). Ensure that the derivative of the antiderivative resembles the original function, indicating correctness.

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Key Concepts

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

Indefinite Integral
An indefinite integral is a fundamental concept in calculus, often described as the "antiderivative" of a function. When we evaluate an indefinite integral, our goal is to find a function whose derivative is equal to the integrand. In simpler terms, we are finding a function that, when differentiated, gives us the function inside the integral sign.
  • Unlike definite integrals, which calculate the exact area under a curve between two points, indefinite integrals represent a family of functions.
  • The result of an indefinite integral includes a constant of integration, typically denoted as "C," which accounts for the vertical shift of the function.
In the exercise provided, you're tasked with evaluating the indefinite integral \( \int xe^{-2x} \, dx \). This means identifying a function whose derivative gives \( xe^{-2x} \), and the solution involves using integration by parts.
Exponential Functions
An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. These functions are characterized by their rapid rate of growth or decay. In this particular exercise, the exponential function present is \( e^{-2x} \).
  • Exponential functions like \( e^x \) and \( e^{-x} \) are unique because their derivative is proportional to the function itself, making them elegant and powerful tools in calculus.
  • The negative sign in \( e^{-2x} \) indicates an exponential decay, slowing the growth of the function as \( x \) increases.
To handle such exponential functions in integration, especially when combined with polynomials like \( x \), the integration by parts technique becomes very useful. It's integral to understand how exponential functions behave to utilize them effectively in calculus problems.
Antiderivative
The antiderivative is essentially the reverse process of differentiation. If you want to think of it as rewinding a function to its original state before differentiation, you're on the right track. Finding the antiderivative is the crux in evaluating indefinite integrals.
  • The antiderivative can be visualized as a curve that results from integrating a given function, along with a constant "C" that accounts for the unknown vertical shift.
  • Antiderivatives are not unique because of this constant of integration; any function of the form \( F(x) + C \) is an antiderivative of \( f(x) \).
In the context of the given exercise, the antiderivative we found for \( xe^{-2x} \) is \( -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C \). This is verified by differentiating the result and checking that it indeed regenerates the original function \( xe^{-2x} \), confirming that we've found the correct antiderivative.

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Most popular questions from this chapter

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^9 \frac{1}{\sqrt[3]{x - 1}}\ dx \)

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(a) Show that \( \displaystyle \int_{-\infty}^\infty x\ dx \) is divergent. (b) Show that $$ \lim_{t\to\infty} \int_{-t}^t x\ dx = 0 $$ This shows that we can't define $$ \int_{-\infty}^\infty f(x)\ dx = \lim_{t\to\infty} \int_{-t}^t f(x)\ dx $$

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^\infty \frac{1}{\sqrt[4]{1 + x}}\ dx \)
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