WILL MAKE YOU THE BRAINEST
Find the area of the region bounded by the curves y equals the inverse sine of x divided by 4, y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.

1/30

Step-by-step explanation:

∫[0,4] arcsin(x/4) dx = 2π-4

x = 4sin(u)

arcsin(x/4) = arcsin(sin(u)) = u

dx = 4cos(u) du

∫[0,4] 4u cos(u) du

∫[0,4] f(x) dx = ∫[0,π/2] g(u) du

v = ∫[1,e] π(R^2-r^2) dx

where R=2 and r=lnx+1

v = ∫[1,e] π(4-(lnx + 1)^2) dx

Using shells dy

v = ∫[0,1] 2πrh dy

where r = y+1 and h=x-1=e^y-1

v = ∫[0,1] 2π(y+1)(e^y-1) dy

v = ∫[0,1] (x-x^2)^2 dx = 1/30

a

step-by-step explanation:

i think the answer is d im not sure