Science:Math Exam Resources/Courses/MATH101/April 2013/Question 12 (a)
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Question 12 (a) 

FullSolution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Prove that diverges. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Intuition should say that the numerator looks like x and the denominator is like and so this function should look like the inverse of x which should diverge. If you're trying to prove it diverges, you don't need to find the exact antiderivative; there are other ways of proving divergence of integrals. 
Hint 2 

Try comparison test. 
Hint 3 

What does the integrand look like for large ? 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Since it is difficult to solve this integral directly, we try the comparison test instead to prove divergence. Notice that for large , the integrand behaves like . Since diverges, it will be our goal to show that our integrand is larger than a constant times . First, since , we have Now, for , we have To prove this, notice if , then where in the second line we added to both sides, and in the third line we divided by . Now, since diverges by ptest, we have that and hence the original integral diverges by the Comparison Test, as required. 