/*! 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 52 Compute the indefinite integral ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 52

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=2^{t} \mathbf{i}+\frac{1}{1+2 t} \mathbf{j}+\ln t \mathbf{k}$$

Short Answer

Expert verified
The indefinite integral of the vector-valued function r(t) with respect to t is: $$\int \mathbf{r}(t) dt = \left(\frac{2^t}{\ln 2} + C_i\right)\mathbf{i} + \left(\frac{1}{2}(\ln|1+2t| + C_j)\right)\mathbf{j} + \left(t \ln t - t + C_k\right)\mathbf{k}$$ where Ci, Cj, and Ck are the constants of integration for the i, j, and k components, respectively.

Step by step solution

01

Find the antiderivative of the i component

To find the antiderivative of the i component (2^t), we can use the standard integration formula for exponential functions: $$\int 2^t dt = \frac{2^t}{\ln 2} + C_i$$ where C_i is the constant of integration for the i component.
02

Find the antiderivative of the j component

To find the antiderivative of the j component (1/(1+2t)), we can use the substitution method. Let u = 1 + 2t, and du = 2dt. Redefine the integral in terms of u: $$\int \frac{1}{1+2t} dt = \frac{1}{2} \int \frac{1}{u} du$$ Now, integrate with respect to u: $$\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2}(\ln|u| + C_j)$$ Substitute u back in terms of t: $$\frac{1}{2}(\ln|1+2t| + C_j)$$
03

Find the antiderivative of the k component

To find the antiderivative of the k component (ln(t)), we can use integration by parts. Let u = ln(t) and dv = dt, then du = (1/t) dt and v = t. $$\int \ln t dt = t \ln t - \int t \frac{1}{t} dt$$ $$t \ln t - \int dt = t \ln t - t + C_k$$
04

Combine the components

Now, we can combine the antiderivatives to form the integral of the vector-valued function: $$\int \mathbf{r}(t) dt = \left(\frac{2^t}{\ln 2} + C_i\right)\mathbf{i} + \left(\frac{1}{2}(\ln|1+2t| + C_j)\right)\mathbf{j} + \left(t \ln t - t + C_k\right)\mathbf{k}$$

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

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Show that \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Express \(\mathbf{I}\) and \(\mathbf{J}\) in terms of the usual unit coordinate vectors i and j. Then, write i and \(\mathbf{j}\) in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

An object moves along a straight line from the point \(P(1,2,4)\) to the point \(Q(-6,8,10)\) a. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with a constant speed over the time interval [0,5] b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{t}\)

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. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)0 \leq t \leq T$$ of an object moving in a gravitational field is a segment of a parabola for some value of \(T>0 .\) Find \(T\) such that \(y(T)=0\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

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.