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Math 2153 Quiz 2, February 2 1. (5 points each) Describe the first step you would take in computing each of the following integrals. Right now our options are “algebraic simplification” (describe the simplification), “guess-and-fix” (describe the initial guess), “u-substitution” (state u and du), “integration by parts” (state u and dv), “use a reduction formula” (write the result of the formula), “trigonometric substitution” (state the rewritten integral), and “use a table” (state the formula you (expect to) find in the table, and the values of any parameters such as u or a). If instead we don’t know a way to replace the integral with a less difficult problem, say that instead. Z √ (a) ( x − 1)2 dx Z √ √ Solution: Do algebra to it. The new integral is ( x)2 − 2( x)(1) + (1)2 dx = Z √ x − 2 x + 1 dx, which is just a polynomial. Z x2 dx √ (b) 6x2 − 49 Solution: Use a trigonometric substitution. 6x2 − 49 looks kind of like sec2 θ − 1 = tan2 θ, so make the substitution 6x2 = 49 sec2 θ, i.e. x = √76 sec θ, which makes dx = √76 sec θ tan θ dθ. The integral becomes Z 49 sec2 θ √7 sec θ tan θ dθ Z 2 6 x dx 6 √ = 49(sec2 θ − 1) 6x2 − 49 Z 49 · 7 · sec2 θ · sec θ tan θ dθ √ = 6 · 6 · 49 tan2 θ Z 7 sec3 θ dθ √ , = 6 6 tan θ which we can finish by using various trigonometric techniques (ask in office hours). Z (c) sec x tan4 x dx 2 2 Solution: Do algebra to it. tan4 x = tan2 x = sec2 x − 1 = Z Z 4 4 2 sec x−2 sec x+1, so sec x tan x dx = sec5 x − 2 sec3 x + sec x dx . We can finish this off using formulas 48 and 13 from the end of the textbook. It isZ disappointing that the textbook doesn’t include a formula for secm u tann u du, but the fact is that it doesn’t, so we have to be able to rewrite integrals like this using other techniques. Z (d) e3x cos 4x dx Z Solution: Use formula 20, The solution is eau cos bu du, with a = 3, b = 4, and u = x. e3x (3 cos 4x + 4 sin 4x) + C. + 42 32 Alternatively, use integration by parts with u = e3x and dv = cos 4x dx or use integration by parts with u = cos 4x and dv = e3x dx . This Z appears to simply transform the problem into e3x sin 4x dx, which is no easier, but then doing it again will yield the formula Z Z 3 3x 9 1 3x 3x e3x cos 4x dx. e cos 4x dx = e sin 4x+ e cos 4x− 4 16 16 Now we can finish the problem using algebra: move the integral 9 over to the left and divide both sides by (1 + 16 ). Extra Credit (3 points): Let C be the part of the curve x2 + y 2 = 4 in the first quadrant. Express the length of C as a definite integral, in a form that a Math 2144 student would understand at the end of the semester. Then evaluate the integral, using techniques from this class. Z end p Solution: The arc-length formula is dx2 + dy 2 . start In our case, the curve starts at (2, 0) and ends at (0, 2) (or, perhaps, goes in the other direction). Differentiating the formula for C, we get 2xdx + 2ydy = d(4), i.e., x 2xdx + 2ydy = 0. So dy = − dx. y Z x=2 s x2 dx2 + 2 dx2 , which still doesn’t make The integral becomes y x=0 sense to a Calculus I student. Z x=2 r x2 2 2 Substituting y = 4 − x , we get dx2 + dx2 , which 4 − x2 x=0 is at least closer. Extracting the common factor of dx2 turns it into Z x=2 s Z x=2 r x2 x2 1+ dx2 = dx , which has the 1+ 2 4−x 4 − x2 x=0 x=0 virtue of looking like integrals are “supposed” to look, and should be recognizable to a student at the end of calculus I.