Question

Compute the average value of the function f ( s ) = sin^3 s cos^8 s

[ 0 , π ]

Answer #2

answered by: anonymous

Compute the average value of the function f ( s ) = sin^5 s
cos^6 son the interval [ 0 , π ]

Identify the surface with parametrization x = 3 cos θ sin φ, y =
3 sin θ sin φ, z = cos φ where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Hint: Find
an equation of the form F(x, y, z) = 0 for this surface by
eliminating θ and φ from the equations above. (b) Calculate a
parametrization for the tangent plane to the surface at (θ, φ) =
(π/3, π/4).

Find the derivative of the following:
f(s)=cos(π/2(s))sin(π/2(s))
If f(q)=e−5qcos(2πq), then f′(q) is
f(y)=cos(π/2*y2)
e3x+7/x2-1
If f(q)=e-5qcos(2πq), then f′(q) is
1/(2−e2s)4=

Find the average of the function ?(?) = sin(?) + cos(2?) over
the interval 0 ≤ ? ≤ 2?.

if f''(x)= -cos(x)+sin(x), and f(0)=1 and f(pi)=), what is the
original function

Find the maximum value of the directional derivative of the
function f(x,y)=cos(3x+2y) at the point (π/6,−π/8). Give an exact
answer.

1. Use the given conditions to find the exact value of the
expression.
sin(α) = -5/3, tan(α) > 0, sin(α - 5π/3)
2. Use the given conditions to find the exact value of the
expression.
cos α = 24/25, sin α < 0, cos(α + π/6)
3. Use the given conditions to find the exact value of the
expression.
cot x = √3, cos x < 0, tan(x + π/6)
4. If α and β are acute angles such that...

1. Find the area between the curve f(x)=sin^3(x)cos^2(x) and y=0
from 0 ≤ x ≤ π
2. Find the surface area of the function f(x)=x^3/6 + 1/2x from
1≤ x ≤ 2 when rotated about the x-axis.

Find f. f ''(θ) = sin(θ) + cos(θ), f(0) = 3, f '(0) = 2

Solve the initial value problem
2(sin(t)dydt+cos(t)y)=cos(t)sin^3(t)
for 0<t<π0<t<π and y(π/2)=13.y(π/2)=13.
Put the problem in standard form.
Then find the integrating factor, ρ(t)=
and finally find y(t)=

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