Find a function f which satisfies the integral equation
Int(bounded from 0 to x) t*f(t)dt = Int(bounded from x to 0)(t^2+1)*f(t)dt +x
To find a function \(f(x)\) that satisfies the given integral equation, we'll solve it step by step.
Step 1: Rewrite the integral equation in terms of definite integrals.
The given integral equation can be rewritten as:
\(\int_{0}^{x} t \cdot f(t) dt = \int_{x}^{0} (t^2 + 1) \cdot f(t) dt + x\)
Step 2: Apply the Second Fundamental Theorem of Calculus.
According to the Second Fundamental Theorem of Calculus, if \(F(x)\) is an antiderivative of \(f(x)\), then:
\(\int_{a}^{b} f(x) dx = F(b) - F(a)\)
So, to solve the integral equation, we need to find the antiderivative \(F(x)\) of \(f(x)\).
Step 3: Rewrite the integral equation using the antiderivative.
Using the Second Fundamental Theorem of Calculus, we can rewrite the integral equation as:
\(F(x) - F(0) = F(0) - F(x) + x\)
Simplifying, we get:
\(2F(x) = 2F(0) + x\)
Step 4: Solve for \(F(x)\).
Dividing both sides by 2, we obtain:
\(F(x) = F(0) + \frac{x}{2}\)
Step 5: Find \(f(x)\).
To find \(f(x)\), we differentiate \(F(x)\) with respect to \(x\):
\(\frac{d}{dx}(F(x)) = \frac{d}{dx}(F(0) + \frac{x}{2})\)
\(f(x) = \frac{1}{2}\)
Therefore, the function \(f(x)\) that satisfies the given integral equation is \(f(x) = \frac{1}{2}\).