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Tugas 3
1.
TUGAS 3 NAMA KELOMPOK
: 1.SUSANDI 2.RAFIS 3.GERIAN KELAS : 1EA LATIHAN 7.1 1. β« 100 ππ₯ = 100π₯ + π 2. β« 6π₯ ππ₯ = 3π₯2 + π 3. β«( 3π₯2 + 4π₯ β 5) ππ₯ = π₯3 + 2π₯ β 5π₯ + π 4. β«( π₯2 + 1)β π₯ ππ₯ = 2 7 π₯2 1 + 2 3 π₯2 3 + π 5. β«( π₯ π + π π₯ ) ππ₯ = π₯ π+1 π+1 + π π₯ + π 6. β«(10π₯ + 30)3 10 ππ₯ = (10π₯+30)4 4 + π 7. β«(π₯2 β 3)4 2π₯ ππ₯ = (π₯2β3)5 5 + π 8. β«(π ππ2 π₯cos π₯) ππ₯ = π ππ3 3 π₯ + π 9. β« π₯2 β π ππ π₯3 ππ₯ = β cos π₯3 3 + π 10. β« πΌπ π₯ ππ₯ = π₯ πΌπ π₯ = π₯ πΌπ π₯ β π₯ + π penyelesaian : 1. = π ππ₯ (100π₯ + π) = 100π₯ + π 2. = π ππ₯ (3π₯2 + π) = 6π₯ + 0 = 100 3. = π ππ₯ (π₯3 + 2π₯2 β 5π₯ + π) = 3π₯2 + 4π₯ β 5 + 0 = 3π₯2 +4x-5 4. = π ππ₯ ( 2π₯ 1 2 7 + 2π₯ 3 2 3 + c)= 2π₯ β1 2 14 + 6π₯ 1 2 6 + 0 = 1π₯ β1 2 7 + π₯ 1 2 5. = π ππ₯ ( π₯ π+1 π+1 + π π₯ + π) = π+1.π₯ π+1β1 1π1β1+0 + π₯π π₯β1 + π = π+π₯ π 1 + π₯π π₯β1 + 0 = π + π₯ π + π₯π π₯β1 6. = π ππ₯ (10π₯+30)4 4 + π = 10π₯4 4 + 304 4 + π = 40π₯3 4 + 0 = 10π₯3 7. = π ππ₯ (π₯2β3)5 5 + c = π₯10 5 β 35 5 + π = 10π₯9 5 + 0 = 2π₯9 8. = π ππ₯ ( π ππ3 3 π₯ + π) = π ππ2 π₯ 3 . sin π₯ 3 + 0 = (1βπππ 2 π₯) . 3 sin π₯ 3 9. = π ππ₯ ( β cos π₯3 3 + π) = β3 sin π₯2 3 + 0 = βsin π₯2 10. = π ππ₯ ( π₯ πΌπ π₯ β π₯ + π) = π₯. ππ₯ π₯ β 1 = π₯β1ππ₯ π₯
2.
LATIHAN 7.2 1. β«8
ππ₯ 2. β« 3 4 ππ₯ 3. β«9.75 ππ₯ 4. β«β3ππ₯ 5. β«( β40 3 β10+15 )ππ₯ 6. β«16 β2 ππ‘ 7. β« π2 ππ₯ 8. β«2π ππ 9. β«β21ππ’ 10.β« 6 π ππ₯ Penyelesaian : 1. = 8x+c 2. = 3 4 π₯ + π 3. = 9π₯. 75π₯ + π 4. = β3 x+c 5. = 40π₯ 2 3 10π₯ 1 2+15π₯ + π 6. = 16π‘. β2 π‘ + c 7. = ππ₯2 + c 8. = 2π. ππ + c 9. = -21 u + c 10.= 6π₯ ππ₯ + c
3.
LATIHAN 7.3 1. β«
π₯5 ππ₯ 2. β« β π₯34 ππ₯ 3. β« π₯β2 ππ₯ 4. β« 1 π₯2 ππ₯ 5. β« π‘100 ππ‘ 6. β« π’2π ππ’ 7. β« 1 βπ₯ ππ₯ 8. β« π₯5 π₯2 ππ₯ 9. β« πβ1 ππ 10.β« 1 π‘ ππ‘ Penyelesaian : 1. = π₯6 6 + π 2. = β« π₯ 3 4 ππ₯ = π₯ 7 4 7 4 + c = 4π₯ 7 4 7 + π 3. = π₯β2+1 β2+1 + π 4. =β« π₯β2 ππ₯ = π₯β1 β1 + π = β 1 π₯ + π 5. = π‘101 101 + π 6. = 42π+1 2π+1 + π 7. = β« π₯ β1 2 ππ₯ = π₯ 1 2 1 2 + π = 2π₯ 1 2 1 + π 8. = π₯6 π₯3 + π 9. = πβ1+1 β1+1 + π = β
4.
10.β« π‘β1 ππ‘ = π‘β1+1 β1+1 +
π = β LATIHAN 7.4 1. β« π π‘ ππ‘ 2. β« π20π₯ ππ₯ 3. β« π ππ₯ ππ₯ 4. β« π0,25π₯ ππ₯ 5. β« π π₯ 5 ππ₯ 6. β« πβ3π₯ ππ₯ 7. β«4 π₯ ππ₯ 8. β«23π₯ ππ₯ 9. β«1000,25π₯ ππ₯ 10.β« π π₯ 5 ππ₯ Penyelesaian : 1. = π π‘ + π 2. = 1 20 π20π₯ + π = π20π₯ 5 + π 3. = 1 π π ππ₯ + π = π ππ₯ π + π 4. = 1 0,25 π0,25π₯ + π = π0,25π₯ 0,25 + π 5. = 1 π₯ 5 π π₯ 5 + π = 5 π₯ π π₯ 5 + π 6. = 1 β3 πβ3π₯ + π = πβ3π₯ β3 + c 7. = 1 πΌπ 4 4 π₯ + π = 4 π₯ πΌπ 4 + π 8. = 1 3 πΌπ 2 23π₯ + π = 23π₯ 3 πΌπ 2 + π 9. = 1 0,25 πΌπ 100 1000,25π₯ + π = 1000,25π₯ 0,25 πΌπ 100 + π 10.= 1 π₯ 5 π π₯ 5 + π = 5 π₯ π π₯ 5 + π
5.
LATIHAN 7.5 1. β«cos
π£ ππ£ 2. β«sin( 1 2 ππ₯)ππ₯ 3. β«cos(18) ππ₯ 4. β« π ππ2 (β3π₯)ππ₯ 5. β« ππ π2 (2,5) ππ₯ 6. β«sec ( 5 6 π₯)tan( 5 6 π₯)ππ₯ 7. β«csc x 3 πππ‘ π₯ 3 ππ₯ 8. β«csc(ex) cot(ππ₯)ππ₯ 9. β«sin 3π ππ 10.β«cos(25ππ₯) ππ₯ Penyelesaian : 1. = Sin v + c 2. = β 1 1 2π cos ( 1 2 ππ₯) + π = β2πcos ( 1 2 ππ₯) + π 3. = 1 8 sin(18π₯) + π 4. = 1 β3 tan(β3π₯) + π = tan(β3π₯) β3 + π 5. = β 1 2,5 cot(2,5 π₯) + π = βcot(2,5π₯) 2,5 + π 6. = 1 5 6 sec( 5 6 π₯) + π = 6 5 sec ( 5 6 π₯) + π 7. = 1 1 3 csc( x 3 ) + c = 3 csc ( π₯ 3 ) + π 8. = β 1 π csc (ex) + c = β csc( ππ₯) π + π 9. = β 1 3 cos(3π) + π = βcos(3π) 3 + c 10.= 1 25π sin(25ππ₯) + π = sin(25 ππ₯) 25π + c
6.
LATIHAN 7.6 1. β« 1 1+π2 ππ 2.
β« ππ₯ β16βπ₯2 3. β« 1 49+π₯2 ππ₯ 4. β« ππ‘ 0,25+π‘2 5. β« ππ’ βπ’2(π’2β1) 6. β« 1 |π₯|βπ₯2β41 ππ₯ 7. β« 1 β 81 100 βπ₯2 ππ₯ 8. β« π2+π₯2 ππ₯ 9. β« ππ‘ βπ‘2(π‘2β 1 4 ) 10.β« 1 |π₯|βπ₯2β7 ππ₯ Penyelesaian : 1. = π‘ππβ1 π + π = βπππ‘β1 π + π 2. = β« 1 β42ββπ₯2 ππ₯ = π ππβ1 ( π₯ 4 ) + π 3. = β« 1 β492+π₯2 ππ₯ = 1 β49 π‘ππβ1 ( π₯ β49 ) + π 4. = β« 1 β0,252+π‘2 ππ‘ = 1 β0,25 π‘ππβ1 ( π‘ β0,25 )+ π 5. = β« 1 |π’|βπ’2ββ12 ππ’ = 1 1 π ππβ1 ( π’ 1 ) + π = π ππβ1( π’) + π 6. = π ππβ1 ( π₯ 41 ) + c = - ππ πβ1 ( π₯ 41 ) + c 7. = β« 1 β (9)2 10 βπ₯2 ππ₯ = π ππβ1 ( π₯ 9 10 ) + π =π ππβ1 ( 10π₯ 9 ) + π = βπππ β1 ( 10π₯ 9 ) + π 8. 1 π π‘ππβ1 π₯ π + π = β 1 π πππ‘β1 ( π₯ π )+c
7.
9.= β« 1 | π‘|β
π‘2β (1) 2 2 ππ₯ = 1 1 2 π ππβ1 ( π‘ 1 2 ) + π = 2π ππβ1(2π‘) + π 10.π ππβ1 ( π₯ 7 ) + π = βππ πβ1 ( π₯ 7 ) + π
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