If I(m, n) = int_0^1 t^m (1 + t)^n dt,then th | vedclass

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(A) We are given $I(m, n) = \int_0^1 t^m (1 + t)^n dt$. Using integration by parts,let $u = (1 + t)^n$ and $dv = t^m dt$. Then $du = n(1 + t)^{n-1} dt

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$\frac{2^n}{m + 1} - \frac{n}{m + 1} I(m + 1, n - 1)$

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(A) We are given $I(m, n) = \int_0^1 t^m (1 + t)^n dt$. Using integration by parts,let $u = (1 + t)^n$ and $dv = t^m dt$. Then $du = n(1 + t)^{n-1} dt$ and $v = \frac{t^{m+1}}{m+1}$. Applying the formula $\int u dv = uv - \int v du$: $I(m, n) = \left[ \frac{t^{m+1}(1 + t)^n}{m+1} \right]_0^1 - \int_0^1 \frac{t^{m+1}}{m+1} \cdot n(1 + t)^{n-1} dt$. Evaluating the boundary terms: At $t=1$,we get $\frac{1^{m+1}(1+1)^n}{m+1} = \frac{2^n}{m+1}$. At $t=0$,the term is $0$. So,$I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} \int_0^1 t^{m+1}(1 + t)^{n-1} dt$. Since $I(m+1, n-1) = \int_0^1 t^{m+1}(1 + t)^{n-1} dt$,we have: $I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1)$.

If $I(m, n) = \int_0^1 t^m (1 + t)^n dt$,then the expression for $I(m, n)$ in terms of $I(m + 1, n - 1)$ is

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