MATH 222 (Lectures,2, and 4) Fall 205 Midterm Solutions Student ID#: Circle your TA s name from the following list. Carolyn Abbott Tejas Bhojraj achary Carter Mohamed Abou Dbai Ed Dewey Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi Chris Janjigian Tao Ju Ahmet Kabakulak Dima Kuzmenko Ethan McCarthy Tung Nguyen Jaeun Park Adrian Tovar Loez Polly Yu Please inform your TA if you find any errors in the solutions. Score Problem Problem 2 Problem Problem 4 Problem 5 Problem 6 Problem 7 Instructions Write neatly on this exam. If you need extra aer, let us know. On Problems, 2, and, only the answer will be graded. On Problems 4, 5, 6, and 7 you must show your work and we will grade the work and your justification, and not just the final answer. Each roblem worth either 4 or 5 oints. No calculators, books, or notes (excet for those notes on your inch by 5 inch notecard.) Please simlify any formula involving a trigonometric function and an inverse trigonometric function. For examle, lease write cos(arcsin x) x 2. Note that we have rovided some formulas on the next age to hel with this.
Formulas You may freely quote any algebraic or trigonometric identity, as well as well as any of the following formulas or minor variants of those formulas. cos(arcsin x) x 2 sec(arctan x) +x 2. tan(arcsec x) x 2. R x n dx ( x n+ n+ R e x dx e x + C R cos xdx sinx + C + C when n 6 ln x + C when n R sin xdx R tan xdx cos x + C ln cos x + C R cot xdx ln sin x + C R sec xdx ln sec x + tan x + C. R csc xdx ln csc x + cot x + C. R +x 2 dx arctan(x)+c.
. For each statement below, CIRCLE true or false. (a) (b) (c) (d) (e) True False True False True False True False True False (a) If x 7 cos then tan 49 x 2 x. (b) R sin 2 ( )d sin cos + C (c) +sin(x) x x for all x. (d) R 2 x 2 9dx is a finite number. (e) R x x x + dx is a finite number. (a) True. (b) False. (c) False. (d) False. (e) True.
2. On this age, only the answer will be graded. (a) Comute R sin 2 (x) cos 2 (x)dx. sin 2 (x) cos 2 (x)x sin 2 (x) ( sin 2 (x))dx (2 sin 2 (x) )dx ( cos(2x)) )dx cos(2x)dx 2 sin(2x)+c (b) Comute R 4 (x )(x+) dx. We rewrite this in the form: 4 (x )(x + ) dx x x + Solving using the method of equating coe cients yields A and B. (c) Comute R x 2 +6x+0 dx. b x 2 +6x + 0 lim b! +(x + ) 2 dx lim b! [arctan(x + )] b lim b! (arctan(b + ) arctan(0)) /2.
. On this age, only the answer will be graded. (a) Find a ositive number A such that R 00 x 2 +7x 5dx < A. Any A bigger than.0075 will work. (b) Comute R xe 7x+ dx. Let f x and g 0 e 7x+ so that f 0 and g 7 e7x+.then xe 7x+ dx fg 0 fg f 0 g x 7 e7x+ 7 e 7x+ dx x 7 e7x+ 49 e7x+ + C (c) Comute R 2x x 2dx. Comlete the square to get 2x x 2 (x ) 2. Then we get: 2x x 2 dx (x ) 2 dx Using x sin and dx cos d this yields: cos d sin 2 d + C arcsin(x ) + C.
4x + 4. Comute dx or exlain why the integral does x(2x +)(2x +) 4x+ not exist. (You may freely use the formula 2x+.) We comute: 4x + x(2x + )(2x + ) dx There are other equivalent answers. lim b! lim b! lim b! x ale ln x " ln ale ln 2x + x(2x+)(2x+) x 2x + dx b 2 ln 2x + 2 ln 2x + x (2x + )(2x + ) # b x 4x 2 +8x + b lim ln b! 4b 2 +8b + ln( 4 ) ln( 5 )ln( b ln 5 5 2 ) 2x+
5. Comute R (z + e z )sin(z)dz. We comute this as the sum of two integrals: (z + e z )sin(z)dz z sin(z)dz + e z sin(z)dz For R z sin(z)dz we double back. First do integration by arts with f z so f 0 and g 0 sin(z) sog cos(z). And we get: z sin(z)dz fg f 0 g z cos(z)+ cos(z)dz z cos(z)+ 9 sin(z)+c Then we let I R e z sin(z)dz. We first integrate by arts with f sin(z) and g 0 e z. Then f 0 cos(z) and g e z, yielding: I e z sin(z)dz fg f 0 g sin(z)e z cos(z)e z dz We integrate by arts again, with h cos(z) and k 0 e z so h 0 sin(z) and k e z : sin(z)e z hk 0 sin(z)e z hk h 0 k sin(z)e z cos(z)e z ( sin(z))e z dz sin(z)e z cos(z)e z 9 sin(z))e z dz sin(z)e z cos(z)e z 9I We thus have the equation: I sin(z)e z cos(z)e z 9I which, after moving all of the I terms to the left side, yields: We thus obtain: ( + 9) I sin(z)e z cos(z)e z + C. I 0 (sin(z)ez cos(z)e z )+C Putting this together yields: (z + e z )sin(z)dz z cos(z)+ 9 sin(z)+ 0 (sin(z)ez cos(z)e z )
6. Comute R e x 4 e 2x dx. Set z e x so that dz e x dx and dx dz e x e x 4 e 2x dx z 2 4 z 2 dz dz z.thenwehave: Now let z 2sin so that dz 2 cos d and we get: (2 sin ) 2 4 4sin 2 2 cos d 4 cos 2 4sin 2 d cos 2 sin 2 d (cot 2 )d csc 2 d cot + C Since sin( ) z 2 we get arcsin( z 2 ) and cot( ) 4 z 2 z yielding: 4 z 2 z 4 e 2x e x arcsin( z 2 )+C arcsin( ex 2 )+C
7. (a) For n 0,,... let I n R x n e x+2 dx. Deriveareductionformula for I n. (b) Let J n R x 5 (ln x) n dx for n 0. This satisfies the reduction formula J n (lnx) n x6 n 6 6 J n for n. Comute J 2. (a): Let f x n so f 0 nx n and let g 0 e x+2 so that g I n x n e x+2 dx fg 0 fg f 0 g xn e x+2 n xn e x+2 n I n x n ex+2 Then: e x+2 dx (b): J 0 R x 5 dx x6 6 Then + C. Then J 2 (lnx) 2 x6 6 J (lnx) x6 6 2 6 J (lnx) 2 x6 6 6 J 0 6 x ln x x 6 6 + C. If you simlify, you might get other, equivalent, answers. 6 x ln x x 6 + C 6