   Chapter 5.3, Problem 58E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# What is wrong with the equation? ∫ 0 π sec 2 x d x = tan x ] 0 π = 0

To determine

To check: The integral function 0πsec2dx=tanx]0π=0, whether it is right or wrong.

Explanation

Given information:

The function is 0πsec2dx=tanx]0π=0 (1)

Apply theorem of fundamental calculus (part 2) as shown in below;

abf(x)dx=F(b)F(a) (2)

Here, F is an anti-derivative of f (F=f).

Modify Equation (1).

abF(x)dx=F(b)F(a)

Check the function whether it is continuous at the interval [0,π].

Substitute 0 for x in Equation (1).

0πsec2x=tan(0)=0

Hence, the value of function is continuous.

Substitute x=π3 in Equation (1).

0πsec2x=tan(π3)=3=1.732

Hence, the value of function is continuous.

Substitute x=π2 in Equation (1) as shown in below;

0πsec2x=tan(π2)=

It is not defined

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