The problem is to evaluate the integral:

$\int_0^1 x^3 \ln(x) \, dx$

Solution:

We'll use integration by parts: Let:

• $u = \ln(x)$, so $du = \frac{1}{x} dx$
• $dv = x^3 dx$, so $v = \frac{x^4}{4}$

Using the formula for integration by parts: $\int u \, dv = uv - \int v \, du$

1. Substituting $u = \ln(x)$ and $v = \frac{x^4}{4}$: $\int x^3 \ln(x) \, dx = \left[ \frac{x^4}{4} \ln(x) \right]_0^1 - \int_0^1 \frac{x^4}{4} \cdot \frac{1}{x} \, dx$

2. First term: At $x = 1$, $\frac{x^4}{4} \ln(x) = \frac{1^4}{4} \ln(1) = 0$ (since $\ln(1) = 0$).

   At $x = 0$, $\frac{x^4}{4} \ln(x)$ is undefined because $\ln(0) \to -\infty$. However, as $x^4 \ln(x) \to 0$ when $x \to 0^+$, this term contributes $0$ (by applying limits).

   So, the first term is $0$.

3. Second term: $\int_0^1 \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{1}{4} \int_0^1 x^3 \, dx$

   Evaluate $\int_0^1 x^3 \, dx$: $\int x^3 \, dx = \frac{x^4}{4} + C$ So: $\left[ \frac{x^4}{4} \right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$

   Thus, the second term becomes: $\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$

Final Answer:

$\int_0^1 x^3 \ln(x) \, dx = -\frac{1}{16}$

Would you like a detailed explanation of any step or have further questions?

Follow-Up Questions:

1. How do you interpret the behavior of $\ln(x)$ at $x = 0$ in integrals?
2. Can you generalize this result for $\int_0^1 x^n \ln(x) \, dx$?
3. What if the integration limits change? How would the calculation differ?
4. How does integration by parts simplify integrals involving logarithms?
5. What are some other applications of integration by parts in physics or engineering?

Tip:

When handling improper integrals, always carefully evaluate the behavior at endpoints to ensure convergence.