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Exercise 1541Combinatorics › Binomial coefficientsL1Added 3 Nov 2026

Application to integration

#integration#binomial#identity

Statement

Compute ∫01(1+x)n dx\int_0^1 (1+x)^n\,dx using two methods.

Hint

Direct: 2n+1−1n+1\frac{2^{n+1}-1}{n+1}. Binomial: integrate the expansion.

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Solution

Direct. ∫01(1+x)n dx=(1+x)n+1n+1∣01=2n+1−1n+1\int_0^1(1+x)^n\,dx=\frac{(1+x)^{n+1}}{n+1}\Big|_0^1=\frac{2^{n+1}-1}{n+1}.

Binomial. ∫01∑k(nk)xk dx=∑k(nk)1k+1\int_0^1\sum_k\binom{n}{k}x^k\,dx=\sum_k\binom{n}{k}\frac{1}{k+1}.

Conclusion. ∑k(nk)k+1=2n+1−1n+1\sum_k\frac{\binom{n}{k}}{k+1}=\frac{2^{n+1}-1}{n+1}.