Calculus/Integration/Solutions

Integration of Polynomials Evaluate the following: 1. 2. 3. 4. 5. Indefinite Integration Find the general antiderivative of the following: 6. 7. 8. 9. 10. 11. 12. Integration by Substitution Find the anti-derivative or compute the integral depending on whether the integral is indefinite or definite. 13. Notice that By this, . Let . Then, Note that a student could do another substitution such as or . The answers should all be the same despite the different methods. We decided to show one method instead of all possible methods. 14. . Rewrite the integral into an equivalent form to help us find the substitution: Let . . Apply all this information to find the original integral: 15. . Let . . Then, 16. . Let . . Then, 17. . It may be easier to see what to substitute once the integrand is written in an equivalent form. From there, it becomes obvious to let . Then, 18. . Let . Then, 19. . Let . Then, . Let . Therefore, Alternatively, this could all be done with one substitution if one realized that . 20. . It may be easier to see what to substitute once the integrand is written in an equivalent form. From there, let . Then, 21. . Let . Then, 22. . Let . Then, This requires another substitution. Let . Then, 23. . Let . Then, Note that the solution of is also correct but not as simplified as it could be. Another thing of note is you can choose your substitution as . If the algebraic steps are correct, your answer should be the exact same as above. Try both methods to get to the same solution. 24. . This integral would be a nightmare to do without first making the form something we can work with. Note the domain of the integrand (which we shall denote as ): , , and (due to the natural domain of the natural logarithm). Solve for . Let . . For all , is strictly increasing. while (show this yourself). By the intermediate value theorem, there must be a solution for for any . Let that solution be . Therefore, the domain for is . Note finding a solution is not possible with standard methods, but can be found using the productlog function. An approximation for is using Newton's method (try it yourself!). Now, attempt to rewrite . This algebraic rewrite is valid if and only if is defined and nonzero. Note however the domain argument from before, where excluding . Since is undefined only at zero, and for all other , . This means the multiplication by on the numerator andn denominator is algebraically valid in its domain. . Then, let Note that the domain must be preserved for the derivative given that the algebraic manipulation may alter the domain. The domain of for all , so . The mapping within the domain of is strictly increasing and one-to-one on . (Notice that the natural log and the rational function is one-to-one. As an exercise to the reader, prove for all ). Therefore, for all in the domain. Meaning the rewritten integrand is valid and does not introduce new singularities. Thus, The domain argument is necessary if a student chooses to rewrite the integral in this way. In fact, this is the only way to integrate if one does integration by substution as traditionally done (i.e., a -sub). This means any student answer that does not confirm nor discusses the domain will not be formally correct! Integration by parts 30. Consider the integral . Find the integral in two different ways. (a) Integrate by parts with and . (b) Integrate by parts with and . Compare your answers. Are they the same? (a) (b) Notice that the answers in parts (a) and (b) are not equal. However, since indefinite integrals include a constant term, we expect that the answers we found will differ by a constant. Indeed Integration by Trigonometric Substitution 40. Let Then