• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 289
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.


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.

See the step by step solution

Step by Step Solution

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.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.