Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.8, Problem 27E

To determine

To calculate: The function $f(x)$ from the equation $f'(t)=2cost+sec2t$ with condition $f(π3)=4,−π2<t<π2$ .

Expert Solution & Answer

Answer to Problem 27E

The function $f(x)=2sint+tan−1t+4−23$ from the equation $f'(t)=2cost+sec2t$ .

Explanation of Solution

Given information:

The equation is given as:

$f'(t)=2cost+sec2t$ and condition $f(π3)=4,−π2<t<π2$ .

Formula used:

The function F is antiderivative of $f(x)$ on an interval $I$ if $F`(x)=f(x)$ for all $x$ in $I$

The function $F$ is known as antiderivative of $f(x)$ on an interval $I$ the most general antiderivative of the function on interval $I$ is $F(x)+C$ , where $C$ is an arbitrary constant.

Calculation:

Consider the function,

$f'(t)=2cost+sec2t$

The antiderivative of $f(x)$ is,

$F`(x)=f(x)$

So, by reverse power rule Integrating both sides of equation,

$∫F`(x)=∫f(x)F(x)=∫f(x)$

Therefore, integrating the function

$∫f'(t)dt=∫2cost+sec2tdtf(t)=2sint+tan−1t+C$

The value of $t$ is alwaysbetween $−π2<t<π2$ because if put $t=π2,t=−π2$ then equation is undefined so value of $−π2<t<π2$

Apply, the condition $f(π3)=4,−π2<x<π2$ on equation for obtain $C$ ,

$f(π3)=2sin(π3)+tan−1(π3)+C4=232+3+C4−23=C$

Put the value of $C$ in the equation,

$f(x)=2sint+tan−1t+4−23$

Thus, the particular solution is $f(x)=2sint+tan−1t+4−23$ .

Chapter 4.8, Problem 26E

Chapter 4.8, Problem 28E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The function f ( x ) from the equation f ' ( t ) = 2 cos t + sec 2 t with condition f ( π 3 ) = 4 , − π 2 &lt; t &lt; π 2 .

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Answer to Problem 27E

Explanation of Solution

Chapter 4 Solutions

The function f ( x ) from the equation f ' ( t ) = 2 cos t + sec 2 t with condition f ( π 3 ) = 4 , − π 2 < t < π 2 .