Cum integrați int x + cosx din [pi / 3, pi / 2]?

Răspuns:

Răspunsul #int _ (pi / 3) ^ (pi / 2) x + cosx * dx = 0,8193637907356557 #

Explicaţie:

arată mai jos

(pi / 3) ^ (pi / 2) x + cosx * dx = 1 / 2x ^ 2 + sinx

# Pi ^ 2/8 + sin (pi / 2) - pi ^ 2/18 + sin (pi / 3) = (5 * pi ^ * 3 ^ 2-4 (5/2) 72) /72=0.8193637907356557#

Răspuns:

#int_ (pi / 3) ^ (pi / 2) (x + cosx) dx = 1 + (5pi ^ 2-36sqrt3)

Explicaţie:

Folosind liniaritatea integrala:

(pi / 3) ^ (pi / 2) (x + cosx) dx = int_ (pi / 3)

Acum:

(pi / 3) ^ (pi / 2) xdx = x ^ 2/2 2) / 72 #

(pi / 3) ^ (pi / 2) cosxdx = sinx (pi / 3) ^ (pi / 2) = sin (pi / #

Atunci:

#int_ (pi / 3) ^ (pi / 2) (x + cosx) dx = 1 + (5pi ^ 2-36sqrt3)

Răspuns:

# (5π ^ 2) / 72 + 1-sqrt3 / 2 #

Explicaţie:

#int_ (π / 3) ^ (π / 2) (x + cosx) dx # #=#

#int_ (π / 3) ^ (π / 2) XDX + int_ (π / 3) ^ (π / 2) cosxdx # #=#

# X ^ 2/2 _ (π / 3) ^ (π / 2) # #+# # Sinx _ (pi / 3) ^ (π / 2) # #=#

# (Π ^ 2/4) / 2- (π ^ 2/9) / 2 + sin (π / 2) -sin (π / 3) # #=#

# π ^ 2/8-π ^ 2/18 + 1-sqrt3 / 2 # #=#

# (5π ^ 2) / 72 + 1-sqrt3 / 2 #