/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Express \(\int x^{2} e^{-x^{2}} ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 19

Express \(\int x^{2} e^{-x^{2}} d x\) in terms of \(\int e^{-x^{2}} d x\).

Short Answer

Expert verified
\( \int x^{2} e^{-x^{2}} \, dx = \frac{1}{2} \int e^{-x^{2}} \, dx \)

Step by step solution

01

Identify the substitution

Use the substitution method. Let us set \[ u = x^2 \] to simplify the integral.
02

Differentiate the substitution

Differentiate the substitution with respect to x:\[ du = 2x \, dx \] or equivalently \[ dx = \frac{du}{2x} \].
03

Modify the integral in terms of u

Substitute back into the integral. Note that \( x^2 = u \) and \( dx = \frac{du}{2x} \):\[\int x^{2} e^{-x^{2}} d x = \int x^2 e^{-u} \frac{du}{2x}.\]
04

Simplify the integral

Simplify the terms:\[\int x^2 e^{-u} \frac{du}{2x} = \int \frac{x^2}{2x} e^{-u} du = \int \frac{x}{2} e^{-u} du = \frac{1}{2} \int u e^{-u} du.\]
05

Recognize the transformed integral

Notice that the term is now in the form of \(\int u e^{-u} du\). To bring it back to the form involving \( \int e^{-x^{2}} d x \), reverse substitute \( u = x^2 \). We then find that\[ \int u e^{-u} du = \int x^2 e^{-x^2} dx.\]
06

Relate to the original integral

Realize the final form is \(= \frac{1}{2} \int u e^{-u} du\) while recognizing what we set for u\(= x^2\), thus confirming\[\int x^{2} e^{-x^{2}} d x = \frac{1}{2} \, \text{with respect to} \, \int e^{-x^{2}} d x \]

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 91影视!

Key Concepts

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

substitution method
Definite integrals are useful for finding the area under a curve within a specified interval. They differ from indefinite integrals by having upper and lower limits of integration. For instance, \( \int_{a}^{b} f(x)dx \) calculates the area between a function \( f(x) \) and the x-axis, from \( x = a \) to \( x = b \).\r\rIn the original problem, though we only dealt with an indefinite integral, understanding the methods for handling definite integrals is essential. For example, if we had specified limits, we would adjust the limits of integration accordingly after substitution. When changing variables, we also have to change the limits of integration to match the new variable. This is crucial for correctly evaluating a definite integral.
techniques of integration
Integration techniques are various methods used to find the integral of a function. These techniques include substitution, integration by parts, partial fractions, and trigonometric integrals, among others. Each method is suitable for different types of functions and can help systematically solve integrals that might otherwise be very difficult to approach.\r\rIn our original problem, we focused on the substitution method. However, it's helpful to know when and how to use other techniques.\r\r
  • Substitution Method: Used for simplifying integrals where a substitution of variables makes the function easier to integrate.
  • Integration by Parts: Suited for integrals that are products of two functions. It uses the formula \( \int uv' = uv - \int u'v \) .\r
  • Partial Fractions: Useful for rational functions, it decomposes a complex fraction into simpler parts before integration.\r
\rBy mastering these techniques, one can tackle a wide range of integrals, simplifying complex problems into manageable steps.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The derivation of \(\int e^{x} \sin x d x\) given in the text seems to prove that the only primitive of \(f(x)=e^{x} \sin x\) is \(F(x)=e^{x}(\sin x-\cos x) / 2,\) whereas \(F(x)=\) \(e^{x}(\sin x-\cos x) / 2+C\) is also a primitive for any number \(C .\) Where does \(C\) come from? (What is the meaning of the equation $$\int e^{x} \sin x d x=e^{x} \sin x-e^{x} \cos x-\int e^{x} \sin x d x ?$$

Suppose that \(f^{\prime \prime}\) is continuous and that $$\int_{0}^{\pi}\left[f(x)+f^{\prime \prime}(x)\right] \sin x d x=2.$$ Given that \(f(\pi)=1,\) compute \(f(0)\).

Potpourri. (No holds barred.) The following integrations involve all the methods of the previous problems. (i) \(\int \frac{\arctan x}{1+x^{2}} d x.\) (ii) \(\int \frac{x \arctan x}{\left(1+x^{2}\right)^{2}} d x.\) (iii) \(\int \log \sqrt{1+x^{2}} d x.\) (iv) \(\int x \log \sqrt{1+x^{2}} d x.\) (v) \(\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{1}{\sqrt{1+x^{4}}} d x.\) (vi) \(\int \arcsin \sqrt{x} d x.\) (vii) \(\int \frac{x}{1+\sin x} d x.\) (viii) \(\int e^{\sin x} \cdot \frac{x \cos ^{3} x-\sin x}{\cos ^{2} x} d x.\) (ix) \(\int \sqrt{\tan x} d x.\) (x) \(\int \frac{d x}{x^{6}+1}.\) (To factor \(x^{6}+1,\) first factor \(y^{3}+1,\) using Problem 1-1.) The following two problems provide still more practice at integration, if you need it (and can bear it). Problem 9 involves algebraic and trigonometric manipulations and integration by parts, while Problem 10 involves substitutions. (Of course, in many cases the resulting integrals will require still further manipulations.)

Find a reduction formula for (a) \(\int x^{n} e^{x} d x\) (b) \(\int(\log x)^{n} d x.\)

The following integrations can all be done with substitutions of the form \(x=\sin u, x=\cos u,\) etc. To do some of these you will need to remember that $$\int \sec x \, d x=\log (\sec x+\tan x)$$ as well as the following formula, which can also be checked by differentiation: $$\int \csc x \, d x=-\log (\csc x+\cot x).$$ In addition, at this point the derivatives of all the trigonometric functions should be kept handy. (i) \(\int \frac{d x}{\sqrt{1-x^{2}}} \cdot\) (You already know this integral, but use the substitution \(x=\sin u\) anyway, just to see how it works out.) (ii) \(\int \frac{d x}{\sqrt{1+x^{2}}}\) (Since \(\tan ^{2} u+1=\sec ^{2} u,\) you want to use the substitution \(x=\tan u .)\) (iii) \(\int \frac{d x}{\sqrt{x^{2}-1}}\). (iv) \(\int \frac{d x}{x \sqrt{x^{2}-1}}\) (The answer will be a certain inverse function that was given short shrift in the text.) (v) \(\int \frac{d x}{x \sqrt{1-x^{2}}}\). (vi) \(\int \frac{d x}{x \sqrt{1+x^{2}}}\). \(\left.\begin{array}{l}\text { (vii) } \int x^{3} \sqrt{1-x^{2}} d x \\ \text { (viii) } \int \sqrt{1-x^{2}} d x\end{array}\right\\} \begin{array}{l}\text { You will need to remember the methods for } \\ \text { integrating powers of sin and cos. }\end{array}.\) (ix) \(\int \sqrt{1+x^{2}} d x\). \((x) \quad \int \sqrt{x^{2}-1} d x\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.