To evaluate the definite integral $\int_0^3 4x^3 \, \text{d}x$, follow these steps:

Step 1: Find the Indefinite Integral

First, find the indefinite integral (antiderivative) of the integrand $4x^3$: $\int 4x^3 \, \text{d}x$ Using the power rule for integration, which states: $\int x^n \, \text{d}x = \frac{x^{n+1}}{n+1} + C$ where $C$ is the constant of integration, we get: $\int 4x^3 \, \text{d}x = 4 \cdot \frac{x^{3+1}}{3+1} = 4 \cdot \frac{x^4}{4} = x^4$

Step 2: Apply the Limits of Integration

Now, evaluate the definite integral by applying the limits from 0 to 3: $\int_0^3 4x^3 \, \text{d}x = \left[ x^4 \right]_0^3$ This means we calculate: $\left[ x^4 \right]_0^3 = (3^4) - (0^4)$

Step 3: Simplify the Result

Calculate the values: $3^4 = 81 \quad \text{and} \quad 0^4 = 0$ Thus: $\int_0^3 4x^3 \, \text{d}x = 81 - 0 = 81$

Final Answer:

$\int_0^3 4x^3 \, \text{d}x = 81$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the result change if the integrand were $4x^2$ instead of $4x^3$?
2. Can you evaluate the definite integral $\int_1^4 x^2 \, \text{d}x$?
3. How do you handle definite integrals where the limits are negative?
4. What is the geometric interpretation of a definite integral?
5. How does the Fundamental Theorem of Calculus connect differentiation and integration?

Tip: Always remember to apply the limits of integration after finding the indefinite integral to evaluate a definite integral.