MAT 213 Chapter 13 HW Solutions.pdf - Solutions Section 13.1 1. ∫ 2. ∫ 3. ∫ 4. ∫ 5. ∫ 6. ∫ 7. 8. 9. ∫ ∫ ∫ 10. 11. 12. 13. 14. = = 6 = 6

1.∫=+1+ 1+;= 5⇒∫5=66+2.∫=+1+ 1+;= 7⇒∫7=88+3.∫6= 6∫1= 6+4.∫(−5)= (−5)∫1=−5+5.∫=+1+ 1+;= 1⇒∫=22+6.∫(−)=−∫=−22+7.∫(2−)=∫2−∫=33−22+8.∫(+3)=∫+∫3=22+44+9.∫(1 +)=∫1+∫=+22+10.∫(4−)=∫4−∫= 4−22+11.∫=+1+ 1+;=−5⇒∫−5=−4(−4)+=−−44+12.∫=+1+ 1+;=−7⇒∫−7=−6(−6)+=−−66+13.∫(2.3+−1.3)=∫2.3+∫−1.3=3.33.3−−0.30.3+14.∫(−0.2−0.2)=∫−0.2−∫0.2=0.80.8−1.21.2+Solutions Section 13.1

15.∫(2−−1)=∫2−∫−1=33−ln | | +16.∫(−2+ 2−1)=−−1+ 2ln | | +=−1+ 2ln | | +17.∫4√=∫1/4=5/45/4+=45/45+18.∫3√=∫1/3=4/34/3+=34/34+19.∫(34−2−2+−5+ 4)= 3∫4−2∫−2+∫−5+∫4=355+ 2−1−−44+ 4+20.∫(47−−3+ 1)= 4∫7−∫−3+∫1=488−−2−2++=82+−22++21.∫ (2+4)=∫(2−1+ (1/4) )= 2ln | | +1422+= 2ln | | +2/8 +22.∫ (22+24)=∫(2−2+ (1/4)2)=2−1−1+1433+=−2−1+3/12 +23.∫ (1+22−13)=∫(−1+ 2−2−−3)= ln | | + 2−1−1−−2−2+= ln | |−2+122+24.∫ (3−15+17)=∫(3−1−−5+−7)= 3ln | |−−4−4+−6−6+= 3ln | | +144−166+25.∫(30.1−4.3−4.1)=31.11.1−5.35.3−4.1+26.∫ (2.12−2.3)=3.16.2−2.3+27.∫ (30.1−41.1)=∫(3−0.1−4−1.1)= 30.90.9−4−0.1−0.1+=0.90.3+400.1+28.(11.1−1)=∫(−1.1−−1)=−0.1−0.1−ln | | +=−100.1−ln | | +Solutions Section 13.1

29.∫(5.1−1.2+31.2)=∫(5.1−1.2−1+ 3−1.2)=5.122−1.2ln | | + 3−0.2−0.2+= 2.552−1.2ln | |−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+ | |/3−1/8)=−4+| |(2)(3)−/8 +=−4+| |6−/8 +33.∫ (6.10.5+0.56−)=∫(6.1−0.5+160.5−)= 6.10.50.5+1.56 · 1.5−+= 12.20.5+1.59−+34.∫ (4.20.4+0.43−2)=∫(4.2−0.4+130.4−2)= 4.20.60.6+1.43 · 1.4−2+= 70.6+1.44.2−2+35.∫(2−3 )=2ln2−3ln3+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(4−324)=∫(2−32−6)=33−32·−5−5+=33+3−510+.40.First multiply inside the integral:∫34(24+356)=∫(6 +95−2)= 6+95·−1−1+= 6−9−15+.41.First decompose the integrand into a sum of two fractions:∫+ 23=∫(3+23)=∫(12+23)=∫(−2+ 2−3)=−−1−−2+=−1−12+.42.First decompose the integrand into a sum of two fractions:∫2−2=∫(2−2)=∫(−2)=∫(−2−1)=22−2ln | | +.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 gives0−0 += 0⇒1 += 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 +=−1⇒2+ 1 +=−1⇒=−2( + 1).So,( ) = 2+−2( + 1).Solutions Section 13.1

Applications47.′( ) = 5−10,000⇒( ) =∫ (5−10,000)= 5−220,000+(0) = 20,000,so0−0 += 20,000,giving= 20,000.Thus,( ) = 5−220,000+ 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+22−−1+(100) = 10,000,so10(100) +10022−1100+= 10,000,giving= 4,000.01.Thus,( ) = 10+22−−1+ 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 = 1,331.775≈1,332million members.