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.

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )