are of a region???? this one is tricky?

Question

The region bounded by the x-axis and the part of the graph of y=sinx between x=0 and x=pi is separated into two regions by the line x=k. If the area of the region for 0<=x<=k ia one-third the area of the region for k<=x<=pi, then k=???? Is it . . .

a) arcsin(1/3)
b) arcsin(1/4)
c) pi/6
d) pi/3
e) pi/4

how did you get this answer????

Answer 1

Ok, let's find first the area between the sin(x) and x - axiss, from x = 0 to x = pi

area = integrate(sinx*dx) x = 0 to x = pi

area = - cosx

-cos(pi) + cos(0) = 1 + 1 = 2

then, the area is separated into two regions, so one region wil be : x = 0 to x = k, and the other region will be : x = k to x = pi.

total area = 2

the area from x = 0 to x = k is 1/2

and the area from x = k to x = pi = 3/2

so, the first area is 1/3 of the second one :

integrate(sinxdx) x = 0 to x = k, = 1/2

-cosk + 1 = 1/2

k = 60º

That's it

Hope that helps

Answer 2

Area1 = integral(0, k) sinx

Area2 = integral(k, pi) sinx

Area2 = 1/3 Area1

Area1= -(cosk - cos0) = -1-cosk
Area2= -(cospi - cosk)=1 + cosk

etc

Answer 3

The two areas are:
[-cos(x)] (0..k) to the left and [-cos(x)](k..pi) to the right.
3(cos(0) - cos(k)) = cos(k) - cos(pi)
3(1 - cos(k)) = cos(k) + 1
3 - 3cos(k) = cos(k) + 1
4cos(k) = 2
cos(k) = 1/2
k = pi/3.
Answer (d).

Answer 4

int (0 to k) sin x dx = 1/3 int (k to π) sin x dx
-cos(k) + cos(0) = 1/3[ -cos(π) + cos(k)]
-cos(k) + 1 = 1/3[ 1 + cos(k)]
-cos(k) + 1 = 1/3 + (1/3)cos(k)
2/3 = (4/3)cos(k)
1/2 = cos k
k = π/3

Answer 5

INT[sin(x)]dx, x=0..k

1/3*INT[sin(x)]dx, x=k..Pi

Integrate each, set equal and solve for k.