Kako integrirate int x + cosx iz [pi / 3, pi / 2]?

Odgovor:

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

Pojasnilo:

pokaži spodaj

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

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

Odgovor:

#int_ (pi / 3) ^ (pi / 2) (x + cosx) dx = 1 + (5pi ^ 2-36sqrt3) / 72 #

Pojasnilo:

Uporaba linearnosti integrala:

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

Zdaj:

#int_ (pi / 3) ^ (pi / 2) xdx = x ^ 2/2 _ (pi / 3) ^ (pi / 2) = pi ^ 2/8-pi ^ 2/18 = (5pi ^ 2) / 72 #

#int_ (pi / 3) ^ (pi / 2) cosxdx = sinx _ (pi / 3) ^ (pi / 2) = sin (pi / 2) -sin (pi / 3) = 1-sqrt3 / 2 #

Nato:

#int_ (pi / 3) ^ (pi / 2) (x + cosx) dx = 1 + (5pi ^ 2-36sqrt3) / 72 #

Odgovor:

# (5π ^ 2) / 72 + 1-sqrt3 / 2 #

Pojasnilo:

#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 #