/*! 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 23 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 23

Evaluate the following integrals. $$\int t^{2} e^{-t} d t$$

Short Answer

Expert verified
Question: Calculate the integral of the function \(t^{2} e^{-t}\) with respect to t. Answer: The integral of the function \(t^{2} e^{-t}\) with respect to t is \(-t^2e^{-t} + 2te^{-t} + 4e^{-t} + C\), where C is the constant of integration.

Step by step solution

01

Recall the integration by parts formula

The integration by parts formula is given by the following equation: $$\int u dv = uv - \int v du$$ We can choose to let \(u=t^{2}\) and \(dv=e^{-t} dt\). To proceed, we need to find du and v.
02

Find du and v

We need to differentiate u with respect to t to find du: $$du = \frac{d}{dt}(t^{2}) = 2t dt$$ Now find the antiderivative of dv: $$v = \int e^{-t} dt = -e^{-t}$$
03

First integration by parts

Now, using the integration by parts formula and the results from steps 2: $$\int t^{2} e^{-t} dt = \left(-t^2e^{-t}\right) \color{white}\underbrace{\!\!-\!\!}_\mathrm{1st\ integration\ by\ parts} \int (-2te^{-t}) dt$$ The remaining integral requires a second application of integration by parts:
04

Setup second integration by parts

Let \(u=2t\) and \(dv=e^{-t} dt\). Next, find du and v: $$du = \frac{d}{dt}(2t) = 2 dt$$ $$v = \int e^{-t} dt = -e^{-t}$$
05

Second integration by parts

Apply the integration by parts formula again: $$\int 2t e^{-t} dt = 2\left[-te^{-t} - \int (-2e^{-t}) dt\right]$$ Now, find the antiderivative of \((-2e^{-t})\): $$\int (-2e^{-t}) dt = 2e^{-t}$$
06

Assemble the final answer

Now, put everything together: $$\int t^{2} e^{-t} dt = -t^2e^{-t} - 2\left[-te^{-t} - 2e^{-t}\right] + C$$ Simplify to get the final answer: $$\int t^{2} e^{-t} dt = -t^2e^{-t} + 2te^{-t} + 4e^{-t} + C$$

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!

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

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 3 x \sin 2 x d x$$

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{x-\sqrt[4]{x}} ; x=u^{4}$$

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Use integration by parts to evaluate the following integrals. $$\int_{0}^{1} x \ln x d x$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

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.