|Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I:||(75)||
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Problem 1. In a very condensed form, the definition of integration is as follows: For bounded on and a partition of set , , and . Then set and . Finally, if we say that `` is integrable on '' and set .
From this definition alone, without using anything proven in class about integration, prove that the function given below is integrable on and compute its integral :
Solution. (Graded by Cristian Ivanescu) Let be an arbitrary partition of . Then for any the infimum is 0 and so . Thus . At the same time, for any the supremum is , and hence and so . Now let be given and let be the partition . Then while and so . Thus . But this is true for any and we already know that . So it must be that . Thus and hence is integrable on and its integral is .
Problem 2. Prove that the function
Solution. (Graded by Julian C.-N. Hung) Differentiate using the first fundamental theorem of calculus. The first summand yields . The second summand is the first summand pre-composed with the function . So by the chain rule, the derivative of the second summand is . is the sum of these two terms, . Hence is a constant.
Problem 3. In class we have proven that a twice-differentiable function satisfying the equation is determined by and . Use this fact and the known formulas for the derivatives of and to derive a formula for in terms of , , and .
Solution. (Graded by Julian C.-N. Hung) Let be a constant and consider the functions and . Then and so both and satisfy . We also have and . So by what we have proven in class or .
Problem 4. The function is defined by .
Solution. (Graded by Vicentiu Tipu)
The results. 82 students took the exam; the average grade was 69.3, the median was 78 and the standard deviation was 26.76.