/*! 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 59 Evaluate the following definite ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 59

Evaluate the following definite integrals. $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the vector function \(\mathbf{i} + t\mathbf{j} + 3t^2\mathbf{k}\) with respect to t from -1 to 1. Answer: The definite integral of the vector function is \(2\mathbf{i} + 2\mathbf{k}\).

Step by step solution

01

Identify the components of the vector function

The given vector function is \(\mathbf{i} + t\mathbf{j} + 3t^2\mathbf{k}\). The i-component is a constant function, 1; the j-component is a linear function, \(t\); and the k-component is a quadratic function, \(3t^2\).
02

Integrate i-component with respect to t

We need to integrate the i-component with respect to t, from -1 to 1. So, $$\int_{-1}^{1} 1 dt$$
03

Evaluate i-component integral

The integral of a constant (1) with respect to t is linear (t). So, $$\left[t\right]_{-1}^{1} = \left[1 - (-1)\right] = 2$$
04

Integrate j-component with respect to t

Now, we need to integrate the j-component with respect to t, from -1 to 1. So, $$\int_{-1}^{1} t dt$$
05

Evaluate j-component integral

The integral of the linear function \(t\) with respect to t is quadratic (\(\frac{1}{2}t^2\)). So, $$\left[\frac{1}{2}t^2\right]_{-1}^{1} = \left[\frac{1}{2}(1^2 - (-1)^2)\right] = \left[\frac{1}{2}(1 - 1)\right] = 0$$
06

Integrate k-component with respect to t

Finally, we need to integrate the k-component with respect to t, from -1 to 1. So, $$\int_{-1}^1 3t^2 dt$$
07

Evaluate k-component integral

The integral of the quadratic function \(3t^2\) with respect to t is cubic (\(t^3\)). So, $$\left[t^3\right]_{-1}^1 = \left[1^3 - (-1)^3\right] = \left[1 - (-1)\right] = 2$$
08

Combine results to form the final vector

Now that we have evaluated the definite integral for each of the components, we can form the resulting vector by combining them: $$\left(2\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}\right) = 2\mathbf{i} + 2\mathbf{k}$$

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

Find the points (if they exist) at which the following planes and curves intersect. $$y+x=0 ; \mathbf{r}(t)=\langle\cos t, \sin t, t\rangle, \text { for } 0 \leq t \leq 4 \pi$$

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the Cauchy-Schwarz Inequality?

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure 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.