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# Solution of Term Exam 3

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Problem 1.

1. Compute .
2. Compute .
3. For , compute .

1. To use the second fundamental theorem of calculus we are looking for a function for which . The most obvious guess is , but this is off by a factor of , for . So a good answer would be . Now

2. Likewise choose to get , and so using the second fundamental theorem of calculus,

3. Let and let . Using the first fundamental theorem of calculus, . So using the chain rule,

Problem 2.

1. Perhaps using L'Hôpital's law, compute and .
2. Use these results to give educated guesses for the values of and (no calculators, please).

1. is differentiable at 0 and . So

(L'Hôpital's law also works and gives the same result).

The second limit is of the form so we can use L'Hôpital:

2. is close to 0, so . Multiplying both sides by we get .

Likewise, , so , so .

Problem 3.

1. State the one partition for every '' criterion of the integrability of a bounded function defined on an interval .
2. Let be an increasing function on and let be the partition defined by , for . Write simple formulas for and for .
3. Under the same conditions, write a very simple formula for .
4. Prove that an increasing function on is integrable.

1. A bounded function defined on an interval is integrable iff for every there is a partition of for which .
2. As is increasing, and . Thus

Likewise, .
3. using telescopic summation this is

4. Since is increasing, is bounded (with upper bound and lower bound ). So using the criterion of part 1, to show that is integrable it is enough to show that for every there is a partition of for which . Indeed, let be given. Choose so big so that , and then the partition of before satisfies , as required.

Problem 4.

1. Show that the function is monotone on the interval .
2. Deduce that for every the equation has a unique solution in the range .
3. For , let be the unique in the range for which . Write a formula for and simplify it as much as you can. Your end result may still contain in it, but not , or .

1. . On we know that , so . So is increasing on .
2. By the theorem about the existence of inverses of monotone functions, has an inverse on and it is defined on . This precisely means that for the equation (which defines ) has a unique solution with in the range .
3. By the theorem about the derivative of an inverse function,

The results. 67 students took the exam; the average grade was 77.7 and the standard deviation was about 22.

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Dror Bar-Natan 2005-02-07