This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 16 1 2 1 dx x How can we extend this to evaluate 1 2 1 dx x ? You guessed it! We will take the limit as the upper limit of integration approaches infinity. That is, = b b dx x dx x 1 2 1 2 1 lim 1 . Evaluate: 1 2 1 dx x Now evaluate 1 1 dx x What makes the difference in the value of the integral? Now evaluate -+ dx e e x x 2 1 The second type of improper integral is the type where there is a point of discontinuity in the interval [a,b]. The discontinuity could occur at either an endpoint or at a point inside the interval. Example: e dx x 2 ln Now evaluate -1 1 2 1 dx x...
View Full Document