Find the area of the region bounded by the curves y equals the inverse sine of x/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

Since f(x) = x3-4x is continuous on [-1,1], the theorem does apply.  

So, by the Mean Value Theorem for integrals, there is a number, c, in the interval (-1,1) such that f(c)(1-(-1)) = ∫(from -1 to 1)f(x)dx              

2f(c) = ∫(from -1 to 1)(x3-4x)dx

2f(c) = [(1/4)x4 - 2x2](from -1 to 1)  

2f(c) = (-1.75 - (-1.75)  

 c3-4c = 0  

 c(c2-4) = 0  

 c = 0, 2, -2  

The only value of c that is in the interval (-1,1) is c = 0.

3000000000000000000000