   Chapter 3.9, Problem 18E

Chapter
Section
Textbook Problem

# Find the most general antiderivative of the function. (Check your answer by differentiation.) g ( v ) = 5 + 3 sec 2 v

To determine

To find:

The most general antiderivative of the given function.

Explanation

1) Concept:

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is,Fx+C where C  is an arbitrary constant.

Definition:

A function F  is called an antiderivative of f on an interval I if

F'x=fx for all x in I.

2) Formula:

The particular antiderivative of sec2x  is tanx

Power rule of antiderivative

ddx xn+1n+1=xn

3) Given:

gv =5+3sec2v

4) Calculation:

The given function is gv =5+3sec2v

The particular antiderivative of sec2x  is tanx

Using the particular antiderivatives,

The antiderivative of the function

gv =5+3sec2v  can be written as

5 v0+10+1+3 (tanv)+C

Which simplifies to,

G

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