Short Answer

What is wrong with the equation?
\(\int\limits_0^\pi {{{\sec }^2}xdx = \left( {\tan x} \right)} _0^\pi = 0\)

Evaluation theorem cannot be applied.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:What’s given.

Given Function is \({\rm{f(x) = se}}{{\rm{c}}^{\rm{2}}}{\rm{xdx}}\)from \({\rm{(\pi ,0)}}\)

Here \(\int\limits_{\rm{0}}^{\rm{\pi }} {} {\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{xdx}}\)does not exist because the function \({\rm{f(\theta ) = se}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\)has an infinite discontinuity at \({\rm{\theta = 0}}\) and \({\rm{\theta = \pi }}\)

Step 2: Checking for Discontinuity.

That is, \({\rm{f}}\)is discontinuous on the interval \({\rm{(\pi ,0)}}\)

Since \({\rm{f}}\) is discontinuous, evaluation theorem cannot be applied.

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Essential Calculus: Early Transcendentals

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Short Answer

What is wrong with the equation?
\(\int\limits_0^\pi {{{\sec }^2}xdx = \left( {\tan x} \right)} _0^\pi = 0\)

Step by Step Solution

Step 1:What’s given.

Step 2: Checking for Discontinuity.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

What is wrong with the equation?\(\int\limits_0^\pi {{{\sec }^2}xdx = \left( {\tan x} \right)} _0^\pi = 0\)

Step by Step Solution

Step 1:What’s given.

Step 2: Checking for Discontinuity.

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What is wrong with the equation?
\(\int\limits_0^\pi {{{\sec }^2}xdx = \left( {\tan x} \right)} _0^\pi = 0\)