Find a formula for the function f(x) such that f′(x)=cos(x)/x and f(2)=3.
cosx/x has no antiderivative in terms of elementary functions.
But using the Ci function,
f(x) = Ci(x) - Ci(2) + 3
Well, let's start with f'(x) = cos(x)/x.
Integrating both sides, we get:
∫[f'(x)] dx = ∫[cos(x)/x] dx
Using integration by parts on the left side, we have:
f(x) = ∫[cos(x)/x] dx = ∫[u dv] dx = uv - ∫[v du]
Where u = 1/x, dv = cos(x) dx, dx = du/x
Applying this to the equation, we have:
f(x) = ∫[cos(x)/x] dx = [1/x * ∫[cos(x)] dx] - ∫[(-1/x^2) * ∫[cos(x)] dx] dx
Simplifying, we get:
f(x) = (1/x) * sin(x) - ∫[(1/x^2) * sin(x)] dx
Now, let's find the antiderivative of -1/x^2 * sin(x):
∫[(1/x^2) * sin(x)] dx = -∫[ (1/x^2) * d(cos(x))] = -∫[ d(cos(x)/x^2)]
This becomes:
∫[d(cos(x)/x^2)] = (cos(x)/x^2) + C
Plugging this back into f(x), we get:
f(x) = (1/x) * sin(x) - (cos(x)/x^2) - C
Now, given f(2) = 3, we can substitute x = 2 and f(x) = 3 into the equation:
3 = (1/2) * sin(2) - (cos(2)/4) - C
Solving for C:
C = (1/2) * sin(2) - (cos(2)/4) - 3
So, the formula for the function f(x) such that f'(x) = cos(x)/x and f(2) = 3 is:
f(x) = (1/x) * sin(x) - (cos(x)/x^2) + [(1/2) * sin(2) - (cos(2)/4) - 3]
To find a formula for the function f(x) such that f'(x) = (cos(x))/x and f(2) = 3, we can solve this as a differential equation problem.
Starting with the given derivative f'(x) = (cos(x))/x, we can integrate both sides:
∫f'(x) dx = ∫(cos(x))/x dx
Using the integral of (cos(x))/x, we need to apply integration by parts. Let's use u = 1/x and dv = cos(x) dx. Then, du = -1/x² dx and v = sin(x).
∫f'(x) dx = sin(x)/x - ∫sin(x)(-1/x²) dx
= sin(x)/x + ∫(sin(x)/x²) dx
Now we have:
f(x) = sin(x)/x + ∫(sin(x)/x²) dx
To evaluate the integral, we use another technique called a Taylor series expansion. The Taylor series expansion for sin(x)/x is:
sin(x)/x = 1 - (x²/3!) + (x⁴/5!) - (x⁶/7!) + ...
So we can rewrite the integral as:
∫(sin(x)/x²) dx = ∫ (1 - (x²/3!) + (x⁴/5!) - (x⁶/7!) + ...) dx
= x - (x³/3*3!) + (x⁵/5*5!) - (x⁷/7*7!) + ...
Therefore, the formula for f(x) is:
f(x) = sin(x)/x + x - (x³/3*3!) + (x⁵/5*5!) - (x⁷/7*7!) + ...
To find the value of the constant C, we can use the condition f(2) = 3.
3 = sin(2)/2 + 2 - (2³/3*3!) + (2⁵/5*5!) - (2⁷/7*7!) + ...
Now, you can calculate the value of the constant C using the Taylor series expansion and the given condition f(2) = 3.
To find a formula for the function f(x) such that f′(x) = cos(x)/x and f(2) = 3, we need to integrate the given derivative function and apply the initial condition.
Step 1: Integrating the derivative function.
The derivative of a function is the rate of change of the function. To find the function itself, we need to integrate its derivative. In this case, we need to integrate f′(x) = cos(x)/x.
∫ f′(x) dx = ∫ (cos(x)/x) dx
To integrate this function, we need to use a standard integral formula or technique. In this case, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's choose:
u = 1/x => du = -1/x² dx
dv = cos(x) dx => v = sin(x)
Using the integration by parts formula, we have:
∫ (cos(x)/x) dx = ∫ u dv
= uv - ∫ v du
= (1/x)(sin(x)) - ∫ sin(x) (-1/x²) dx
= (sin(x)/x) + ∫ (sin(x)/x²) dx
Now, we need to integrate ∫ (sin(x)/x²) dx. This is a well-known integral that can be solved using the formula ∫ (sin(x)/x) dx = Si(x), where Si(x) is the sine integral function.
Therefore:
∫ (cos(x)/x) dx = (sin(x)/x) + Si(x) + C
Step 2: Apply the initial condition.
We are given f(2) = 3. Substituting x = 2 into the function we derived in step 1:
f(2) = (sin(2)/2) + Si(2) + C = 3
Now, we need to solve for C:
(sin(2)/2) + Si(2) + C = 3
C = 3 - (sin(2)/2) - Si(2)
Therefore, the formula for the function f(x) is:
f(x) = (sin(x)/x) + Si(x) + (3 - (sin(2)/2) - Si(2))
Note: Si(x) is a special function, the sine integral, and it does not simplify further.