1. TESTS FOR CONVERGENCE AND DIVERGENCE Section 8.3b

2. Sometimes we cannot evaluate an improper integral directly  In these cases, we first try to determine whether it converges or diverges. Diverges???... End of story, we’re done. Converges???... Use numerical methods to Approximate the value of the integral.

3. Investigating Convergence Does the integral converge? Definition from last class: We cannot directly evaluate this last integral directly because there is no simply formula for the antiderivative of the integrand… Instead, compare the integral to one we can evaluate… Graph both functions in [0, 3] by [–0.5, 1.5]: And note that for

4. Investigating Convergence Does the integral converge? Thus, for any Because the integral of the larger function converges, the integral of the smaller function converges as well. However, this does not tell us the value of the improper integral, except that it is positive and less than 0.368…

5. Investigating Convergence Does the integral converge? Evaluate numerically: Try graphing: in [0, 20] by [–0.1, 0.3] What does the graph suggest as ?

6. Direct Comparison Test Let and be continuous on and let for all. Then 1. converges if converges. 2. diverges if diverges.

7. Investigating Convergence Does the integral converge? Compare this integrand to : on So because the integral of this “larger” function converges, the original integral converges as well!

8. Investigating Convergence Does the integral converge? Compare this integrand to : on So because the integral of this “smaller” function diverges, the original integral diverges as well!

9. Limit Comparison Test If the positive functions and are continuous on and if thenand both converge or diverge If two functions grow at the same rate as x approaches infinity (recall Section 8.2?), then their integrals from a to infinity behave alike; they both converge or both diverge.

10. Investigating Convergence Show that converges by comparison with Find and compare the two integral values. The two functions are positive and continuous on the given interval… Use the new rule: Form Thus, both integrals converge. Now to find their value…

11. Investigating Convergence Show that converges by comparison with Find and compare the two integral values. (from earlier in class)

12. Investigating Convergence Show that converges. Compare with:  This integral converges (an example from earlier in class) to a value of about 0.368… Thus, the original integral also converges!

13. Investigating Convergence Use integration, the direct comparison test, or the limit comparison test to determine whether the given integral converges or diverges. The integral converges

14. Investigating Convergence Use integration, the direct comparison test, or the limit comparison test to determine whether the given integral converges or diverges. Compare to : on Since this integral diverges, the given integral diverges