Let's solve the problem step by step.

Given Integral Equation:

$\int_0^{x^2} t f(t) \, dt = \sin(x) - x\cos(x)$

Step 1: Differentiate Both Sides with Respect to $x$

We use the Leibniz rule for differentiation under the integral sign:

$\frac{d}{dx} \int_{a(x)}^{b(x)} F(t) dt = F(b(x)) \cdot b'(x) - F(a(x)) \cdot a'(x)$

Since the lower bound is $0$, its derivative is zero. The upper bound is $x^2$, so its derivative is:

$\frac{d}{dx} \int_0^{x^2} t f(t) \, dt = (x^2 f(x^2)) \cdot \frac{d}{dx} (x^2) = x^2 f(x^2) \cdot 2x = 2x^3 f(x^2).$

On the right-hand side, differentiate term by term:

$\frac{d}{dx} [\sin(x) - x \cos(x)] = \cos(x) - \left( \cos(x) - x \sin(x) \right) = x \sin(x).$

Thus, we obtain:

$2x^3 f(x^2) = x \sin(x).$

Step 2: Solve for $f(x)$

Substituting $u = x^2$, so $x = \sqrt{u}$, we rewrite:

$2 u^{3/2} f(u) = \sqrt{u} \sin(\sqrt{u}).$

Dividing both sides by $2u^{3/2}$, we get:

$f(u) = \frac{\sin(\sqrt{u})}{2u}.$

Since $u = x^2$, we write:

$f(x) = \frac{\sin(x)}{2x}.$

Part (a): Compute $f(1)$

$f(1) = \frac{\sin(1)}{2(1)} = \frac{\sin(1)}{2}.$

Part (b): Compute $f'(1)$

Using the quotient rule for $f(x) = \frac{\sin(x)}{2x}$:

$f'(x) = \frac{(2x \cos(x)) - (2 \sin(x))}{4x^2}.$

$f'(x) = \frac{2x\cos(x) - 2\sin(x)}{4x^2} = \frac{x\cos(x) - \sin(x)}{2x^2}.$

Evaluating at $x = 1$:

$f'(1) = \frac{1\cos(1) - \sin(1)}{2(1)^2} = \frac{\cos(1) - \sin(1)}{2}.$

Final Answers:

(a) $f(1) = \frac{\sin(1)}{2}$.

(b) $f'(1) = \frac{\cos(1) - \sin(1)}{2}$.

Would you like a more detailed explanation of any step?

Here are some related questions:

1. How does the Leibniz rule for differentiation under the integral sign work?
2. Can we use the Fundamental Theorem of Calculus here?
3. How do we check if $f(x)$ is continuous for all $x > 0$?
4. What happens if we change the upper bound to $x^3$ instead of $x^2$?
5. Can we generalize this method for other integral equations?

Tip:

Always verify whether a function found through differentiation satisfies the original equation by substituting it back.