Kako ocenjujete določen integralni int t sqrt (t ^ 2 + 1dt), ki ga omejuje [0, sqrt7]?

Kako ocenjujete določen integralni int t sqrt (t ^ 2 + 1dt), ki ga omejuje [0, sqrt7]?
Anonim

je

# int_0 ^ sqrt7 t * sqrt (t ^ 2 + 1) dt = int_0 ^ sqrt7 1/2 * (t ^ 2 + 1) '* sqrt (t ^ 2 + 1) dt = int_0 ^ sqrt7 1/2 * (t ^ 2 + 1) ^ (3/2) / (3/2) 'dt = 1/3 * (t ^ 2 + 1) ^ (3/2) _ 0 ^ sqrt7 = 1/3 (16 sqrt (2) -1) ~~ 7.2091 #

Odgovor:

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) dt = 7.209138999 #

Pojasnilo:

Iz danih

#int tsqrt (t ^ 2 + 1) "" dt = # omejeno # 0, sqrt7 #

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) dt = 1 / 2int_0 ^ sqrt7 2t (t ^ 2 + 1) ^ (1/2) "" dt #

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) "" dt = 1/3 * (t ^ 2 + 1) ^ (3/2) # od 0 do # sqrt7 #

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) dt = 1/3 (sqrt7 ^ 2 + 1) ^ (3/2) - (0 ^ 2 + 1) ^ (3/2) #

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) "" dt = 1/3 (8) ^ (3/2) - (+ 1) ^ (3/2) #

# int_0 ^ sqrt7 tsqrt (t ^ 2 + 1) dt = 7.209138999 #

Bog blagoslovi … Upam, da je razlaga koristna.