# 1ab - MA 241 Test 1 Form E Spnng 9 Set 11 ONLY a b and c...

This preview shows pages 1–8. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA 241 Test 1 Form E ) Spnng 9. Set 11 ONLY a b and c below. Insert all numerical values so that if you had a calculator, you could ﬁnish the arithmetic. a. Use Simpson’s Rule with n=4 subintervals to approximate 3 x f—dx 1 1 + (sin x)2 b. Find the area of the region R bounded by the line y=x and the parabola 2 y=6—x_ c. Find the volume of the solid generated by revolving the region in part b (previous area problem) about the line y=10. 10. Decompose to partial fractions (this is not asking for integration!) x (x + 3)(x — 2)2 Continue on the next page Form B continued 11. Integrate. x a. ——dx fx+l b. I coshxg + 3) dx 6. f xzedex d. fxln(5x) dx x-2 . ————dx 6 f(x2-4x+l)2 12. Determine if the improper integral converges or diverges. If it converges, evaluate. °° l f 2 dx 0 x + 4 13. Use the appropriate formula, if needed, from the given ones list below from the Tables of Integrals to integrate x e f 2x dx 3—6 . Table of Integrals: du l u+a a. = —ln +C faz —u2 2a u—a du l u—a b. = —ln +C fuz —a2 2a u+a Fab .9 200 _ b ﬂag» l, a u L V . AVI 9U’Xz 1+ x” ' :0 L x+37 x-2)= 0 >002 ' «l0 . v . . C v",' ' 3' .0. M0 l l. (A. 224’ M ‘ ’ ‘ ' ’ AWL“ ML?- , o 2 2 ~ R'dx '.“'_.n -. A ;.‘._,~‘,-4.'_aum:z~ v...” u -' -* ~ Fawn E, p31 - ‘- ; X’Z + x+’\$ X-z + 4?; — =A- ~22+-3+3)—*2+z/-+3 -:- 7-.» o— +2 . Pom E, 733 j, I“ I"; L i ' MA 241 Spring 2007 Test 1 solution key 1 1 (a) Integrate a: /\$+1dx Solution: This fraction is improper, hence we perform division. ___.__1 as + 1) a: —m—1 — 1 Therefore :1: _ 1 _ 1 x+1— x+1 And we now integrate ll x 1 /x+1d\$ _/(1_x+1>dx flair—f 1 d9: x+1 m—ln|m+1|+C H 11(b) Integrate \/_ cos( x + 3) / —\/E da: Solution: Use u—substitution. 1 u=\/§+3, duzmdcc Integrate: cos(\/E + 3) _ cosﬁ/E + 3) = 2/cos(\/§+3)—}——dx 2% 2 / cos(u)du 2 sin(u) + C 2 sin(\/§ + 3) + 0 Pink version _____ __<_A___T_.,.=A,_1 ._‘_..A‘. -«1- __......_._.4_ ._..._V 11(c) Integrate 1. E? 2 x3 an 6 dx i! Solution: Use u—substitution l u = x3,du = 3x2da: /:I:26\$3d117 % f BmZexsda: % / eudu 1 = geu + C = \$623 + C 11(d) Integrate / m1n(5:c)d:c Solution: Use integration by parts (LIATE) 1 5 1 i u = ln(5x), du = 5—mdx = de I 2 a: du — a: (1512,12 — —2— f x ln(5m)d:1: judo uv—fvdu 2 21 a: 1n(5x)_/x__dm H i 2 2 :1: x2 1n(533) a: _ 2 - 5d”: l = x2 1n(5:c) _ it: + C, [\D 4; 1 1 (e) Integrate / a: — 2 dm (3:2 —- 450 + 1)2 Solution: Use u—substitution u=w2—4z+1,du=(2x—4)dx H l / 23—2 d 1/ 293—4 dm (x2—4x+1)2 m 2 (m2—4x+1)2 _1du _2u2 H ml b—l g| w 9.. Q lu—l — 5:” 1 — —%+C — — 1 +0 _ 2(x2—4m+1) 12 Determine of the improper integral converges or diverges. If it converges, evaluate. °° 1 d [0 x2+4\$ Solution: °° 1 t 1 = 1' f0 \$2+4dx tiglo O x2+4dx _ 1 t x) m=t _ t—Ig2arc an<2 :c=0 — 11 1 t t 1 t (0) — Ergo 2 are an 2 2 are an . 1 t = tljgloéarctan (§)—0 _ 177 _ 22 _ E _ 4 3 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern