# MAT 213 Chapter 13 HW Solutions.pdf - Solutions Section...

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1.=+1+ 1+;= 55=66+2.=+1+ 1+;= 77=88+3.6= 61= 6+4.(5)= (5)1=5+5.=+1+ 1+;= 1=22+6.()==22+7.(2)=2=3322+8.(+3)=+3=22+44+9.(1 +)=1+=+22+10.(4)=4= 422+11.=+1+ 1+;=55=4(4)+=44+12.=+1+ 1+;=77=6(6)+=66+13.(2.3+1.3)=2.3+1.3=3.33.30.30.3+14.(0.20.2)=0.20.2=0.80.81.21.2+Solutions Section 13.1
15.(21)=21=33ln | | +16.(2+ 21)=1+ 2ln | | +=1+ 2ln | | +17.4=1/4=5/45/4+=45/45+18.3=1/3=4/34/3+=34/34+19.(3422+5+ 4)= 3422+5+4=355+ 2144+ 4+20.(473+ 1)= 473+1=48822++=82+22++21.∫ (2+4)=(21+ (1/4) )= 2ln | | +1422+= 2ln | | +2/8 +22.∫ (22+24)=(22+ (1/4)2)=211+1433+=21+3/12 +23.∫ (1+2213)=(1+ 223)= ln | | + 21122+= ln | |2+122+24.∫ (315+17)=(315+7)= 3ln | |44+66+= 3ln | | +144166+25.(30.14.34.1)=31.11.15.35.34.1+26.∫ (2.122.3)=3.16.22.3+27.∫ (30.141.1)=(30.141.1)= 30.90.940.10.1+=0.90.3+400.1+28.(11.11)=(1.11)=0.10.1ln | | +=100.1ln | | +Solutions Section 13.1
29.(5.11.2+31.2)=(5.11.21+ 31.2)=5.1221.2ln | | + 30.20.2+= 2.5521.2ln | |150.2+30.(3.2 +10.9+1.23)=(3.2 +0.9+1.2/3)= 3.2 +0.10.1+2.23(2.2)+= 3.2 + 100.1+2.26.6+31.(2+ 5| | + 1/4)= 2+5 | |2+4+32.(4+ | |/31/8)=4+| |(2)(3)/8 +=4+| |6/8 +33.∫ (6.10.5+0.56)=(6.10.5+160.5)= 6.10.50.5+1.56 · 1.5+= 12.20.5+1.59+34.∫ (4.20.4+0.432)=(4.20.4+130.42)= 4.20.60.6+1.43 · 1.42+= 70.6+1.44.22+35.(23 )=2ln23ln3+36.(1.1+ 2 )=1.1ln1.1+2ln2+37.∫ (100(1.1 )2| |3)=(100(1.1 )23| |)=100(1.1 )ln(1.1)23·| |2+=100(1.1 )ln(1.1)| |3+38.∫ (1, 000(0.9 ) +4| |5)=(1, 000(0.9 ) +45| |)=1, 000(0.9 )ln(0.9)+45·| |2+=1, 000(0.9 )ln(0.9)+2 | |5+39.First multiply inside the integral:Solutions Section 13.1
2(4324)=(2326)=3332·55+=33+3510+.40.First multiply inside the integral:34(24+356)=(6 +952)= 6+95·11+= 6915+.41.First decompose the integrand into a sum of two fractions:+ 23=(3+23)=(12+23)=(2+ 23)=12+=112+.42.First decompose the integrand into a sum of two fractions:22=(22)=(2)=(21)=222ln | | +.43.( ) =,so( ) ==22+.To find the value ofwe use the given information that(0) = 1.Substituting gives022+= 1= 1.So,( ) =22+ 1.44.( ) =1,so( ) =1= ln | | +.To find the value ofwe use the given information that(1) = 1.Substituting givesln1 += 1= 1.So,( ) = ln | | + 1.45.( ) =1,so( ) =(1)=+.To find the value ofwe use the given information that(0) = 0.Substituting gives00 += 01 += 0=1.So,( ) =1.46.( ) = 2+ 1,so( ) =(2+ 1)= 2++.To find the value ofwe use the given informationthat(1) =1.Substituting gives21+ 1 +=12+ 1 +=1=2( + 1).So,( ) = 2+2( + 1).Solutions Section 13.1
Applications47.( ) = 510,000( ) =∫ (510,000)= 5220,000+(0) = 20,000,so00 += 20,000,giving= 20,000.Thus,( ) = 5220,000+ 20,000.48.( ) = 10 +2100,000( ) =∫ (10 +2100,000)= 10+3300,000+(0) = 100,000,so0 + 0 += 100,000,giving= 100,000.Thus,( ) = 10+3300,000+ 100,000.49.( ) = 5 + 2+1( ) =(5 + 2+1)= 5+2+ ln+(Note that> 0,so there is no need to writeln | |.)(1) = 1,000,so5 + 1 + ln1 += 1,000,giving= 994.Thus,( ) = 5+2+ ln+ 994.50.( ) = 10 ++12( ) =(10 ++12)= 10+221+(100) = 10,000,so10(100) +100221100+= 10,000,giving= 4,000.01.Thus,( ) = 10+221+ 4,000.01.51. a.We are given( ) =( ) =6.62+ 18 + 220,so( ) =(6.62+ 18 + 220)=2.23+ 92+ 220 +.To findwe use the fact that(0) = 360.Substituting gives360 =2.2(0)3+ 9(0)2+ 220(0) +=,so( ) =2.23+ 92+ 220 + 360.b.Midway through 2014 (= 4.5), the membership was(4.5) =2.2(4.5)3+ 9(4.5)2+ 220(4.5) + 360 = 1,331.7751,332million members.

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