MT171 - FEEDBACK on QUESTION SHEET 7 - 2018/19 MARKING COMMENTS Questions marked are: Q2/Q3(b) and (e)/ Q4 (a)/ Q4(d) /Q5(b)and(d) Each question will be classified as 0-if it is mostly incorrect; 1-if it is about half correct; 2-if it is mostly correct. The maximum you can get is 10. Comment 1 (C1) You need to improve the clarity of your presentation. Comment 2 (C2) Explain what you are doing. Comment 3 (C3) Use correct notation! An equation needs to be written here. This needs a LHS (left hand side) and a RHS (right hand side)! Comment 4 (C4) Use correct mathematical language/notation. Comment 5 (C5) You didn't complete question 5 so there is no Mathematica notebook attached: that is why you have a 'zero' in Continuous assessment. Comment 6 (C6) State your conclusion! Comment 7 (C7) Needs more details on what you are doing! Comment 8 (C8) Where is your integration element? An integral can't be written without it. Q2 Comment 2.1 (C2.1) Your work is correct but if you take the logarithmic derivative first and after differentiate, you get the result almost immediately (less time!). In exams you are also evaluated on the time you take to solve problems so it is important to use the new techniques to calculate efficiently your result! Comment 2.2 (C2.2) Getting to the correct result with incorrect steps will not give you the marks: it is better to use correct steps and arrive to a consistent result (even if that is not the required one!) .... Comment 2.3 (C2.3) You need to show the individual derivatives calculated before you put them in the Leibniz formula: those details are needed for you to get the marks!
Q3 (b) and (e) Comment 3.1 (C3.1) Again the most efficient way to deal with this simple integral is by showing that the integrand function is odd, i.e., f(-x)=(1+x)(-x)(1-x) =- f (x) and hence its integral over a symmetric interval with respect to the origin is zero! We proved this result in lectures. The procedure again gives you more time in exams and sometimes (like in part (d)) makes your life easier. Comment 3.2 (C3.2) You need to evaluate two integrals with different functional forms in different intervals because of the modulus function x -1 = (x-1) for x>1 x -1 = (1-x) for x<1 You can't ignore the modulus function. Please see solutions. Q4 (a, d) Comment 4.1 (C4.1) You need to state that your constant, C for example, is an arbitrary constant of integration. Comment 4.2 (C4.2) Do not forget to calculate the derivative of the variables of integration: if the first variable is t and the substitution is u you have to evaluate either dt du ?du or ?dt , depending on what you find more convenient. Otherwise you get an incorrect integral and all your efforts are not rewarded! Comment 4.3 (C4.3) When you substitute the integration variable the end limit of the integral has to be substituted too. Comment 4.4 (C4.4) When