Definite Integration Question - None

Question

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

Accepted Answer

Solution: Use integration by parts on $I(m, n) = \int_0^1 t^m(1+t)^n dt$.

Let $u = (1+t)^n$ and $dv = t^m dt$, so $du = n(1+t)^{n-1}dt$ and $v = \frac{t^{m+1}}{m+1}$.

$$I(m,n) = \left[\frac{t^{m+1}}{m+1}(1+t)^n\right]_0^1 - \int_0^1 \frac{t^{m+1}}{m+1} \cdot n(1+t)^{n-1}dt$$

$$= \frac{2^n}{m+1} - \frac{n}{m+1}\int_0^1 t^{m+1}(1+t)^{n-1}dt$$

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

∴ Answer is (A).

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