(c) The volume produced by revolving the area enclosed by the first arch of y = the line y = 4 sin(x) and the x-axis, around (d) Find the area between y = sin(x) and y = ACCURATE drawing of this area. cos(x) on the interval 0

int) Approximate cos(4.6) using a quadratic approximation: First note that cos(4.6) cos(3/2). Let f(x) = cos(x). Then, f'(x) = -sinx and f"(x) = -cosx Let a = 3/2. Then f' (3/2) = -sin(3pi/2) and f" (31/2) = -cos(3pi/2) Q(x), the quadratic approximation to cos(x) at a = 3π/2 is: Q(x) = Use Q(x) to approximate cos(4.6). cos(4.6) last 2

2.5 cm/s. H what rate' E. The lateral area A d3 right circular cone a radis rand height h i gilen b a height The radiu Sin Creasing at The rate 48 cm/s, while The heightis at the rate of. 2.5 cm/s. H what rate' is The bater al area a The cone In creasing of Sangin g when r=GO cm and h=80 cm. Express your answer n tems g Ti.

Find dy/dx

Find y'. Given: y = -sin(4x)cot(4x) %3D

Integral 2. secEx dx tanx

Find the exact area inside the curve defined by r = 2 + 3 sin 0 and outside the circle r = 2.

Let y = cot (csc2x/x3+5). Find y'

The area of the shaded region under the curve y = 2 sin 2x in (a) equals the area of the shaded region under the curve y = sin x in (b). Explain why this is true without computing areas. y= 2 sin 2x y sin x (a) (b) Why do both regions have the same area? Choose the correct answer below. A. The area under a sine wave is always 2. Therefore, both curves have the same area. B. When the change of variables u = 2x is applied to the definite integral 2 sin 2x dx, the new integral evaluates to the same value as sin x dx from 0 to x. c. When the chain rule u = 2x is applied to the indefinite integral 2 sin 2x dx, the new integral evaluates to the same value as sin x dx from 0 to . D. When the change of variables u = 2x is applied to the definite integral sin x dx, the nev integral evaluates to the same value as 2 sin 2x dx, from 0 to z.