Question: User: $\int\sum_{i=1}^ni^2$
1. **Stating the problem:**
We want to understand and evaluate the expression involving the integral and the summation: $$\int \sum_{i=1}^n i^2$$.
2. **Understanding the summation:**
The summation $$\sum_{i=1}^n i^2$$ means adding the squares of integers from 1 to $$n$$.
3. **Formula for the sum of squares:**
The formula for the sum of the first $$n$$ squares is:
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$
4. **Interpreting the integral:**
The integral $$\int \sum_{i=1}^n i^2$$ is ambiguous without limits or a variable of integration. Usually, summations are discrete and integrals are continuous operations.
5. **If the integral is with respect to $$n$$ treated as a continuous variable:**
We replace the summation by its formula and integrate:
$$\int \frac{n(n+1)(2n+1)}{6} \, dn$$
6. **Expanding the integrand:**
$$\frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 3n^2 + n}{6}$$
7. **Integral becomes:**
$$\int \frac{2n^3 + 3n^2 + n}{6} \, dn = \frac{1}{6} \int (2n^3 + 3n^2 + n) \, dn$$
8. **Integrate term-by-term:**
$$\frac{1}{6} \left( \frac{2n^4}{4} + \frac{3n^3}{3} + \frac{n^2}{2} \right) + C = \frac{1}{6} \left( \frac{n^4}{2} + n^3 + \frac{n^2}{2} \right) + C$$
9. **Simplify:**
$$= \frac{n^4}{12} + \frac{n^3}{6} + \frac{n^2}{12} + C$$
10. **Final answer:**
$$\int \sum_{i=1}^n i^2 \, dn = \frac{n^4}{12} + \frac{n^3}{6} + \frac{n^2}{12} + C$$
**Note:** The integral is indefinite and assumes $$n$$ is continuous. If the integral or summation context differs, please clarify.