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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.

Suppose f is continuous and that for all x > 0, the integral from 0 to x^2 of t*f(t) dt equals sin(x) - x*cos(x). Compute f(1) and f'(1).

Learn how to compute f(1) and f'(1) given an integral equation involving t*f(t), trigonometric functions, and the Leibniz rule for differentiation under the integral sign. Includes step-by-step solutions using advanced calculus techniques.

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