find the area in the first quadrant bounded above by y=sinx, below by the x-axis, and to the right by x=pi/2 is divided into two equal parts by the line x=a. Find a.
To find the value of "a" that divides the area in the first quadrant into two equal parts, we need to set up an integral for the given area and solve it.
The area in the first quadrant bounded above by y = sin(x), below by the x-axis, and to the right by x = π/2 can be represented using the definite integral:
A = ∫[a, π/2] [sin(x)] dx
Since we want to divide the area into two equal parts, we can split the integral into two equal parts as follows:
A/2 = ∫[a, π/2] [sin(x)] dx / 2
To solve this equation, we need to evaluate both sides of the equation:
∫[a, π/2] [sin(x)] dx / 2 = ∫[0, π/2] [sin(x)] dx - ∫[0, a] [sin(x)] dx
Now we can evaluate the integrals on the right side of the equation:
∫[a, π/2] [sin(x)] dx / 2 = [-cos(x)] from 0 to π/2 - [-cos(x)] from 0 to a
Plugging in the limits of integration:
∫[a, π/2] [sin(x)] dx / 2 = [-cos(π/2) + cos(0)] - [-cos(a) + cos(0)]
Simplifying:
∫[a, π/2] [sin(x)] dx / 2 = [0 + 1] - [-cos(a) + 1]
∫[a, π/2] [sin(x)] dx / 2 = 1 + cos(a)
Now we have the equation:
1 + cos(a) = ∫[a, π/2] [sin(x)] dx / 2
Since we want to divide the area into two equal parts, we set A/2 equal to 1 + cos(a).
A/2 = 1 + cos(a)
To find the value of "a" that satisfies this equation, we need to solve for "a".