Solve 3cosx-1,0<=x<=pi/2

E(-9)≈3.4936

Step-by-step explanation:

Calculate the arc length of the following curve from x = 0 to x = π/2:

y(x) = -1 + 3 cos(x)

Hint: | The definition of arc length in Cartesian coordinates is s = integral_(x_0)^(x_1) sqrt(1 + y'(x)^2) dx.

Apply the definition of arc length to y(x) = -1 + 3 cos(x) for 0<x<π/2:

s = integral_0^(π/2) sqrt(1 + (d/dx(-1 + 3 cos(x)))^2) dx

Hint: | What is d/dx(-1 + 3 cos(x))?

Compute the derivative d/dx(-1 + 3 cos(x)):

= integral_0^(π/2) sqrt(1 + (-3 sin(x))^2) dx

Hint: | Can the integrand be simplified?

Simplify sqrt(1 + (-3 sin(x))^2):

= integral_0^(π/2) sqrt(1 + 9 sin^2(x)) dx

Hint: | Can this integral be computed?

Compute the definite integral:

E(-9)≈3.4936

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes