calculus help por favor?

zackary · May 2021

let r be the region in the first quadrant enclosed by the graphs of f(x)=8x^3 and g(x)= sin(pie*x).

(A) write an equation for the line tangent to the graph of f at x=1/2

(B) find the area of r

(C) write, but do not evaluate and integral expression for the volume of the solid generated when r is rotated about the horizontal line y=1.

ironboxy · May 2021

a.

at x = 1/2, f(1/2) = 8(1/2)^3 = 8(1/8) = 1

f(x) = 8x^3

f'(x) = 24x^2

f'(1/2) = 24(1/4) = 6

y - 1 = 6(x - (1/2))

b.

f, and g's intersection is at x = -1/2, and x = 1/2.

area r = area on the right of the y axis.

area r = ∫[0, 1/2] (sin(πx) - 8x^3)dx

= [ -(1/π)cos(πx) - 2x^4 ] | [0, 1/2]

= [ -(1/π)cos(π/2) - 2(1/16)] - [-(1/π)cos(0) - 2(0)]

= [ 0 - (1/8) ] - [ -(1/π) - 0]

= (-1/8) - (-1/π)

= 1/π - 1/8

c.

volume = ∫π(outer radius^2 - inner radius^2)dx

= π∫[0, 1/2] ((1 - 8x^3)^2 - (1 - sin(πx))^2)dx

^ write but do not evaluate, so leave it like that.

incase you're curious, here's the actual value.

Hope this helps

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