To solve the problem and determine $f(1)$, we proceed as follows:

We are given the equation: $\int_0^x f(t) \, dt + \int_0^{x^2} t f(t) \, dt = 3x^3$ where $f$ is a continuous function. The goal is to find $f(1)$.

Step 1: Differentiate both sides with respect to $x$

To eliminate the integrals, differentiate the equation with respect to $x$ using the Fundamental Theorem of Calculus and the Chain Rule.

1. For the first term $\int_0^x f(t) \, dt$, the derivative is: $\frac{d}{dx} \left( \int_0^x f(t) \, dt \right) = f(x).$

2. For the second term $\int_0^{x^2} t f(t) \, dt$, use the chain rule. Let $u = x^2$, so: $\frac{d}{dx} \left( \int_0^{x^2} t f(t) \, dt \right) = \frac{d}{du} \left( \int_0^u t f(t) \, dt \right) \cdot \frac{du}{dx}.$ By the Fundamental Theorem of Calculus: \frac{d}{du} \left( \int_0^u t f(t) \, dt \right) = u f(u) \quad \text{(since the upper limit is u)}. Therefore: $\frac{d}{dx} \left( \int_0^{x^2} t f(t) \, dt \right) = (x^2) f(x^2) \cdot 2x = 2x^3 f(x^2).$

3. The derivative of the right-hand side $3x^3$ is: $\frac{d}{dx} (3x^3) = 9x^2.$

Step 2: Combine the derivatives

Now combine the results of the differentiation: $f(x) + 2x^3 f(x^2) = 9x^2.$

Step 3: Set $x = 1$

Substitute $x = 1$ into the equation to find $f(1)$.

1. $f(1) + 2(1)^3 f(1^2) = 9(1)^2$: $f(1) + 2 f(1) = 9.$

2. Combine like terms: $3f(1) = 9.$

3. Solve for $f(1)$: $f(1) = 3.$

Final Answer:

$f(1) = 3.$

Do you need further clarification on any step? Let me know!

Here are 5 related questions for practice:

1. Evaluate $f(x)$ for any other $x$ if the same equation holds.
2. Differentiate and solve an equation involving nested integrals like $\int_0^x \int_0^t f(s) \, ds \, dt$.
3. Solve for $f(x)$ if the equation involves multiple integrals with trigonometric limits.
4. Apply the Fundamental Theorem of Calculus to solve integrals with variable limits.
5. Determine the function $f(x)$ if given $\int_0^x t^n f(t) \, dt = Cx^{n+1}$ where $C$ is a constant.

Tip: Always remember to use the Fundamental Theorem of Calculus and chain rule when differentiating integrals with variable upper limits.