Math 222, Exam I, September 17, 2002 Answers

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1 Math, Exam I, September 7, 00 Answers I. (5 points.) (a) Evaluate (6x 5 x 4 7x + 3/x 5 + 4e x + 7 x ). Answer: (6x 5 x 4 7x + 3/x 5 + 4e x + 7 x ) = = x 6 + x 3 3 7x + 3 ln x 5x + 4ex + 7x ln 7 + C. Answer: t 6 t dt. t 4 t 6 t t 4 dt = (t t ) dt = ( ) t t = ( ) ( ) 3 + Overall, students did very well on this. About 5% of the students made one or more of these mistakes: a) people thought the integral of 7 x is 7 x+ /(x + ). They just need to memorize this. a) some people thought you just had one big fraction as the integrand. I guess we can mention this to them, but as we said, it was on the homework. b) some people did not see to split it up into two fractions, and hence were unable to do the problem. People tried to factor the numerator didn t get anywhere. II. (5 points.) Evaluate cos x. + cos x Answer: cos x = = x sin x + + C. 4 sin x cos x Answer: The substitution u = sin x, du = cos x gives sin x cos x = u du = u3 3 + C = sin3 x + C. 3

2 The most common mistakes on problem were when students did not see the easy substitution. They tried integration by parts and reduction formulas, and most often these methods didn t work. cos x was missed far more often than sin x cos x. There were also a few students who mismemorized the formula: cos x = ( + cosx)/. III. (5 points.) Evaluate sin(3x) cos(5x). Answer: From the formula sin(α ± β) = sin α cos β ± cos α sin β we get sin 8x = sin(5x + 3x) = sin 5x cos 3x + cos 5x sin 3x so Hence sin x = sin(5x 3x) = sin 5x cos 3x cos 5x sin 3x sin(3x) cos(5x) = sin 3x cos 5x = cos 5x sin 3x = sin 8x sin 8x sin x. cos 8x cos x sin x = + + C. 6 4 (i) There were about 3 people who had no idea what to do here (because they forgot the relevant trig identity). (ii) There were quite a few people who remembered the identity sin(x + y) = sin(x) cos(y) + sin(y) cos(y) but could not remember/derive the identity sin(x y) = sin(x) cos(y) sin(y) cos(x). These people would obtain the original integral in terms of the integral for sin(5x) cos(3x) and could not go any further. (iii) Of the people who did remember both identities, some could not obtain a correct formula for sin(3x) cos(5x): they would subtract the two lines instead of adding them, or would actually obtain a formula for sin(5x) cos(3x) and (incorrectly) use that formula instead of the formula for sin(3x) cos(5x). (It turns out that the difference is only a minus sign, so they were not too far off). (iv) Of the people who did get the identity for sin(3x) cos(5x) correctly, many (this was the most common error) integrated the sine or the cosine incorrectly, messing up the signs.

3 (v) About a fourth of the students tried integration by parts instead of using the trig identities, and only about 3 people managed to do it correctly. The problem CAN be done using integration by parts twice. (v) About two thirds of the people who tried integration by parts ended up making the wrong substitution the second time around, basically UNdoing their first integration by parts. Only one person noticed that they had undone their first integration by parts (and obtained something like A = B B +A). Most people who undid their integration by parts did sign errors on their integration of sine and cosine and/or on the coefficients of the integration to obtain something like A = B + B + sa (where s was some fraction), which allowed them to find an (incorrect) answer for A. (vi) Of the people who did not undo their integration by parts the second time, most did sign errors in their integration of sine and cosine and/or committed errors on the coefficients of the sine cosines terms (something like integrating cos(3x) got them a 3 multiplying sin(3x) instead of a third multiplying sin(3x)). Nonetheless, these people had the right idea, and got an expression like A = B + sa, which allowed them to find an expression for A which was incorrect only because of their sign/coefficient errors. IV. (5 points.) Evaluate 4 x. Answer: Use the substitution x = sin θ, θ = sin (x/), cos θ = 4 x, = cos θ dθ. Then cos θ ( x ) = = dθ = θ + C = sin + C. 4 x cos θ V. (5 points.) (a) Evaluate x + 6x + 0. Answer: Completing the square gives x +6x+0 = (x+3) + which suggests the substitution x + 3 = tan θ,x + 6x + 0 = sec θ, = sec θ dθ. Thus sec x + 6x + 0 = θ dθ = θ + c = tan (x + 3) + C. sec θ x + 6x + 8. Answer: Completing the square gives x + 6x + 8 = (x + 3) which suggests either the substitution x+3 = sec θ,x +6x+8 = tan θ, = tan θ sec θ dθ or the substitution x+3 = cosh t,

4 x + 6x + 8 = sinh t, = sinh t dt. Although both these methods work, it is easiest to use partial fractions as follows. so x + 6x + 8 = (x + 3) = ((x + 3) + )((x + 3) ) = (x + 4)(x + ) x + 6x + 8 = (x + 4)(x + ) = A x B x + = / x / x + by the Heaviside trick and so x + 6x + 8 = ln x ln x + + C. Because the two problems appeared together under the heading of completing the square and because I told the students when you see x + a try x = a cos θ and when you see x a try x = a sec θ many of the students used the substitution x + 3 = sec θ instead of the method of partial fractions. I will rewrite this question so as to indicate that the method of partial fractions is easier. Another point is that students left answers in a form like sin(cos (x)) rather than simplifying to get x. I did not deduct points for this because of my general policy: Unless otherwise stated, a formula which gives the correct answer when evaluated by a simple (nonsymbolic) calculator is acceptable. It might be wise to find a wording for this question which forces the simplification. (I dislike the instruction Simplify your answer because it is vague.) VI. (5 points.) (a) Evaluate x sin x. Answer: Using Integration by parts with u = x, v = cos x, du =, dv = sin x we get x sin x = u dv = uv v du = x cos x + cos x = x cos x + sin x + C. x cos x. Answer: Using Integration by parts with u = x, v = sin x, du =, dv = cos x

5 we get x cos x = u dv = uv v du = x sin x x sin x. Hence by the previous problem x cos x = x sin x ( x cos x + cos x + C ). Students generally did pretty well on this problem. Almost everyone knew that they were supposed to use integration by parts, and tried to apply a formula for it that was at least close to the correct one. The most common mistakes were:. Taking u = sin x instead of u = x, and ending up with an integral with a higher power of x than the one they started with. (Maybe 30% did this). Taking the derivative of the wrong part at the wrong time. Some students would mess this up even if they that the formula written down correctly. (Maybe 0% did this) 3. Of the ones that did everything else correctly, about / (or a little less, probably) lost track of their minus signs on the way to their final answer. Thu Sep 9 09:5: 00 There are 78 scores grade range count percent A % AB % B % BC % C % D % F % Mean score = 7.. Mean grade =