1 3 ∫ ( √𝑡 − 2)𝑑𝑡 −1 1 1 3 ∫ √𝑡 𝑑𝑡 − ∫ 2 𝑑𝑡 −1 1 −1 1 1 ∫ 𝑡 3 𝑑𝑡 − 2 ∫ 𝑑𝑡 −1 −1 𝑡=1 1 1 +1 3 1 ( +1) 𝑡 3 − 2𝑡| 𝑡=−1 𝑡=1 3 (4 ) 𝑡 3 − 2𝑡| 4 𝑡=−1 4 4 3 3 ( ) ( ) ( (1) 3 − 2(1)) − ( (−1) 3 − 2(−1)) 4 4 3 3 ( − 2) − ( + 2) 4 4 3 8 3 8 ( − )−( + ) 4 4 4 4 3−8 3+8 ( )−( ) 4 4 5 11 16 − − =− = −4 4 4 4 𝑦 = 4 − 𝑥2 𝑛 𝑏 ∫ 𝑓(𝑥) 𝑑𝑥 = lim ∑ 𝑓(𝑎 + 𝑘∆𝑥)∆𝑥 𝑎 𝑛→∞ 𝑘=1 𝑎 = −2 ; 𝑏 = 2 ; 𝑓(𝑥) = 4 − 𝑥 2 ; ∆𝑥 = 𝑓(𝑎 + 𝑘∆𝑥) = 𝑓 (−2 + 𝑓(𝑥) = 4 − 𝑥 2 → 𝑏 − 𝑎 2 − (−2) 4 = = 𝑛 𝑛 𝑛 4𝑘 ) 𝑛 𝑓 (−2 + 2 4𝑘 4𝑘 ) = 4 − ( − 2) 𝑛 𝑛 4𝑘 2 4𝑘 = 4 − [( ) − 2 ( ) (2) + 22 ] 𝑛 𝑛 16𝑘 2 16𝑘 =4−[ 2 − + 4] 𝑛 𝑛 =4− 16𝑘 2 16𝑘 + −4 𝑛2 𝑛 𝑓(𝑎 + 𝑘∆𝑥) = − 16𝑘 2 16𝑘 + 𝑛2 𝑛 𝑛 ∑ 𝑓(𝑎 + 𝑘∆𝑥)∆𝑥 = (− 𝑘=1 16𝑘 2 16𝑘 4 + )( ) 𝑛2 𝑛 𝑛 𝑛 𝑛 𝑛 𝑘=1 𝑘=1 𝑘=1 𝑛 𝑛 𝑛 𝑘=1 𝑘=1 64𝑘 2 64𝑘 64𝑘 2 64𝑘 ∑ (− 3 + 2 ) = ∑ (− 3 ) + ∑ ( 2 ) 𝑛 𝑛 𝑛 𝑛 64𝑘 2 64𝑘 64𝑘 64𝑘 2 ∑ (− 3 + 2 ) = ∑ ( 2 ) − ∑ ( 3 ) 𝑛 𝑛 𝑛 𝑛 𝑘=1 = 𝑛 𝑛 𝑘=1 𝑘=1 64 64 ∑ 𝑘 − 3 ∑ 𝑘2 2 𝑛 𝑛 = 64 𝑛(𝑛 + 1) 64 𝑛(𝑛 + 1)(2𝑛 + 1) ( )− 3( ) 2 𝑛 2 𝑛 6 = 32(𝑛 + 1) 32(𝑛 + 1)(2𝑛 + 1) − 𝑛 3𝑛2 = 32𝑛 + 32 64𝑛2 + 96𝑛 + 32 − 𝑛 3𝑛2 = 32𝑛 + 32 64𝑛2 + 96𝑛 + 32 − 𝑛 3𝑛2 = 32𝑛 32 64𝑛2 96𝑛 32 + − − 2− 2 𝑛 𝑛 3𝑛2 3𝑛 3𝑛 = 32 + 32 64 96 32 − − − 𝑛 3 3𝑛 3𝑛2 lim (32 + 𝑛→∞ 32 64 96 32 64 32 − − − 2 ) = 32 − = = 10,667 𝑛 3 3𝑛 3𝑛 3 3 2 2 2 ∫ 4 − 𝑥 2 𝑑𝑥 = ∫ 4 𝑑𝑥 − ∫ 𝑥 2 𝑑𝑥 −2 2 −2 −2 2 2 𝑥3 ∫ 4 𝑑𝑥 − ∫ 𝑥 𝑑𝑥 = 4𝑥 − | 3 −2 −2 −2 2 (4(2) − (2)3 (−2)3 ) ) − (4(−2) − 3 3 8 −8 32 (8 − ) − (−8 − )= = 10.66 3 3 3