Chapter 7: Problem 24
In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{0}^{3} x^{2} e^{-x} d x $$
Short Answer
Expert verified
The integral evaluates to approximately \( 2 - \frac{5}{e^{3}} \).
Step by step solution
01
Identify Functions for Integration by Parts
In integration by parts, we choose two functions based on the integral \( \int u \, dv = u v - \int v \, du \). For \( u \), we select a function that simplifies upon differentiation and for \( dv \), a function easy to integrate. Here, let \( u = x^2 \) and \( dv = e^{-x} \, dx \).
02
Differentiate and Integrate
Differentiate \( u \): \( du = 2x \, dx \). Now integrate \( dv \): \( \int e^{-x} \, dx = -e^{-x} \), so \( v = -e^{-x} \).
03
Apply the Integration by Parts formula
Plug \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int u \, dv = u v - \int v \, du = x^2(-e^{-x}) - \int (-e^{-x})(2x) \, dx \].
04
Simplify the Calculation
Compute \( x^2 (-e^{-x}) \) to get \( -x^2 e^{-x} \). Then, calculate the integral \( \int 2x e^{-x} \, dx \).
05
Repeat Integration by Parts for Second Integral
For \( \int 2x e^{-x} \, dx \), set \( u = 2x \) and \( dv = e^{-x} \, dx \). Then \( du = 2 \, dx \) and \( v = -e^{-x} \). Repeat the process: \[ 2x(-e^{-x}) + \int 2(-e^{-x}) \, dx = -2x e^{-x} - 2 \int e^{-x} \, dx \].
06
Evaluate Remaining Integral
For \( \int e^{-x} \, dx \), previously calculated as \(-e^{-x}\). Substitute back to find: \( -2x e^{-x} + 2e^{-x} \).
07
Combine All Parts
Combine results: \[ -x^2 e^{-x} - (-2x e^{-x} + 2e^{-x}) \]. This simplifies to \(-x^2 e^{-x} + 2x e^{-x} - 2e^{-x} \).
08
Evaluate Definite Integral from 0 to 3
Evaluate at bounds 3 and 0: \[ \left[-x^2 e^{-x} + 2x e^{-x} - 2e^{-x} \right]_0^3 \]. Calculate separately at 3 and 0: \((-(9)e^{-3} + 6e^{-3} - 2e^{-3}) - (0 + 0 - 2)\).
09
Simplify and Arrive at Solution
Simplify calculations: \([-5e^{-3}] - [-2]\). That results in \( -5e^{-3} + 2 \). Numerical value using \( e \approx 2.71828 \) can be further simplified if required.
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.
Definite Integrals
Definite integrals are a fundamental concept in calculus that help us determine the total accumulation of quantities. In simple terms, it calculates the area under a curve from one point to another. This is particularly useful in various fields, such as physics and engineering, where understanding the net accumulation or total change is crucial.
To calculate a definite integral, we need to evaluate the antiderivative at two bounds. In our exercise, the bounds are from 0 to 3. This process involves not only finding an indefinite integral but carefully substituting these boundaries to find the final result.
The definite integral formula is expressed as:
In our problem, this step was crucial to find the area under the curve \( x^{2} e^{-x} \) from 0 to 3.
To calculate a definite integral, we need to evaluate the antiderivative at two bounds. In our exercise, the bounds are from 0 to 3. This process involves not only finding an indefinite integral but carefully substituting these boundaries to find the final result.
The definite integral formula is expressed as:
- \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \\]
In our problem, this step was crucial to find the area under the curve \( x^{2} e^{-x} \) from 0 to 3.
Differentiation
Differentiation is another core principle of calculus, and it plays a significant role in evaluating integrals, especially when using integration by parts. Differentiation is essentially finding the rate at which a function changes at any point.
When using integration by parts, we start with a function that simplifies upon differentiation. In the step-by-step solution, differentiation was used to find \( du \) from our selected \( u = x^2 \). By differentiating, we determined that \( du = 2x \, dx \).
Understanding how to differentiate a function correctly is crucial when performing integration by parts, as it allows us to simplify complex integrals into manageable parts.
When using integration by parts, we start with a function that simplifies upon differentiation. In the step-by-step solution, differentiation was used to find \( du \) from our selected \( u = x^2 \). By differentiating, we determined that \( du = 2x \, dx \).
Understanding how to differentiate a function correctly is crucial when performing integration by parts, as it allows us to simplify complex integrals into manageable parts.
- The derivative of \( x^n \) is \( nx^{n-1} \). So, for \( x^2 \), it becomes \( 2x \).
Indefinite Integrals
An indefinite integral is essentially the reverse of differentiation. It represents a family of functions that can produce a given derivative. Indefinite integrals come without specified limits and always include an integration constant \( C \).
Within the integration by parts process, finding indefinite integrals is necessary. For instance, finding \( v \) involves integrating \( dv \). Starting from \( dv = e^{-x} \, dx \), we find that \( v = -e^{-x} \). This integral provides a function without specific bounds, which we later use within the bounds to find definite integrals.
Indefinite integrals are an essential step in the process, whereby:
Within the integration by parts process, finding indefinite integrals is necessary. For instance, finding \( v \) involves integrating \( dv \). Starting from \( dv = e^{-x} \, dx \), we find that \( v = -e^{-x} \). This integral provides a function without specific bounds, which we later use within the bounds to find definite integrals.
Indefinite integrals are an essential step in the process, whereby:
- They are solved using basic integration rules, such as \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \).