Given f''(x)= -4sin(2x) and f'(0)=4 and f(0) =-6. Find f(pi/2)
f" = -4sin2x
f' = 2cos2x + c
2+c = 4
c=2
so, f' = 2cos2x + 2
f = sin2x + 2x + c
0+0+c = -6
c = -6
so, f = sin2x + 2x - 6
To find f(pi/2), we need to integrate f''(x) twice and apply the initial conditions f'(0)=4 and f(0)=-6.
First, let's integrate f''(x) with respect to x to get f'(x):
∫(f''(x) dx) = ∫(-4sin(2x) dx)
Using the double angle identity for sine, sin(2x) = 2sin(x)cos(x), we have:
-4∫(2sin(x)cos(x) dx)
Applying the power reduction formula, which states that sin(2x) = 2sin(x)sqrt(1-sin^2(x)), we have:
-8∫(sin(x)cos(x) dx)
Using the integration formula ∫(sin(x)cos(x) dx) = (sin^2(x))/2 + C1:
-8((sin^2(x))/2 + C1)
Simplifying further:
-4sin^2(x) - 8C1
Now, let's integrate f'(x) with respect to x to get f(x):
∫(f'(x) dx) = ∫(-4sin^2(x) - 8C1 dx)
Using the power reduction formula ∫(sin^2(x) dx) = (x - sin(x)cos(x))/2 + C2, we have:
-4((x - sin(x)cos(x))/2 + C2) - 8C1
Simplifying further:
-2x + 2sin(x)cos(x) + 8C2 - 8C1
Now, we can apply the initial conditions to find the values of the constants C1 and C2.
Given f'(0) = 4, we substitute x=0 into the expression for f'(x):
-2(0) + 2sin(0)cos(0) + 8C2 - 8C1 = 4
Simplifying further:
8C2 - 8C1 = 4
Given f(0) = -6, we substitute x=0 into the expression for f(x):
-2(0) + 2sin(0)cos(0) + 8C2 - 8C1 = -6
Simplifying further:
8C2 - 8C1 = -6
We now have a system of equations:
8C2 - 8C1 = 4
8C2 - 8C1 = -6
Subtracting the second equation from the first equation, we obtain:
0 = 10
This equation has no solution, which means there is no unique solution for f(x) based on the given information. Therefore, we cannot determine the value of f(pi/2) from the given conditions.
To find f(pi/2), we need to integrate f''(x) twice and apply the given initial conditions.
Step 1: Integrate f''(x) to find f'(x).
∫f''(x)dx = ∫(-4sin(2x))dx
f'(x) = -4∫sin(2x)dx
Using the integration formula for sin(x): ∫sin(x)dx = -cos(x),
we get:
f'(x) = -4(-1/2cos(2x)) = 2cos(2x) + C₁,
where C₁ is the constant of integration.
Step 2: Apply the initial condition f'(0) = 4 to find C₁.
f'(0) = 2cos(2(0)) + C₁
4 = 2cos(0) + C₁
C₁ = 4 - 2
C₁ = 2
So, the equation for f'(x) is:
f'(x) = 2cos(2x) + 2.
Step 3: Integrate f'(x) to find f(x).
∫f'(x)dx = ∫(2cos(2x) + 2)dx
f(x) = 2∫cos(2x)dx + 2∫dx
Using the integration formula for cos(x): ∫cos(x)dx = sin(x),
we get:
f(x) = 2(1/2sin(2x)) + 2x + C₂
f(x) = sin(2x) + 2x + C₂,
where C₂ is the constant of integration.
Step 4: Apply the initial condition f(0) = -6 to find C₂.
f(0) = sin(2(0)) + 2(0) + C₂
-6 = 0 + 0 + C₂
C₂ = -6
So, the equation for f(x) is:
f(x) = sin(2x) + 2x - 6.
Step 5: Plug in x = pi/2 to find f(pi/2).
f(pi/2) = sin(2(pi/2)) + 2(pi/2) - 6
f(pi/2) = sin(pi) + pi - 6
f(pi/2) = 0 + pi - 6
f(pi/2) = pi - 6.
Therefore, f(pi/2) = pi - 6.