If I_m,n = \int\limits_0^1 x^m - 1(1 - x)^n - 1dx , for m, n \ge 1 , and \int\limits_0^1 x^m - 1 + x^n - 1 \over (1 + x)^m + 1 dx = \alpha I_m,n\alpha \in R , then \alpha equals \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q45 with Solution

View written solutionFree

Correct answer: 1

\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$$ for $m,n\ge 1$. We need to evaluate $$J=\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}\,dx$$ (since the intended denominator is $(1+x)^{m+n}$ so that the result is proportional to $I_{m,n}$), and find $\alpha$ such that $$J=\alpha I_{m,n}.$$ 2. Split the integral: $$J=\int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}\,dx+\int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}\,dx.$$ Let $$J_1=\int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}\,dx.$$ Use the substitution $$x=\frac{t}{1-t}, \qquad t=\frac{x}{1+x}, \qquad dx=\frac{dt}{(1-t)^2}.$$ When $x=0$, $t=0$; when $x=1$, $t=\frac12$. Also, $$x^{m-1}=\left(\frac{t}{1-t}\right)^{m-1}, \qquad (1+x)^{m+n}=\left(\frac1{1-t}\right)^{m+n}.$$ Thus, $$J_1=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt.$$ Similarly, $$J_2=\int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}\,dx=\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt.$$ Hence, $$J=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt+\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt.$$ 3. In the second integral, put $t=1-u$. Then when $t=0$, $u=1$, and when $t=\frac12$, $u=\frac12$. So, $$\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt =\int_{1/2}^1 u^{m-1}(1-u)^{n-1}\,du.$$ Therefore, $$J=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt+\int_{1/2}^1 t^{m-1}(1-t)^{n-1}\,dt.$$ So, $$J=\int_0^1 t^{m-1}(1-t)^{n-1}\,dt=I_{m,n}.$$ Thus, $$\alpha=1.$$ 4. Final answer: $$\boxed{1}$$ 5. Comparison with stored correct answer: Stored correct answer = $1$. Our derived answer matches it.

View written solutionFree

Correct answer: 1

\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$$ for $m,n\ge 1$. We need to evaluate $$J=\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}\,dx$$ (since the intended denominator is $(1+x)^{m+n}$ so that the result is proportional to $I_{m,n}$), and find $\alpha$ such that $$J=\alpha I_{m,n}.$$ 2. Split the integral: $$J=\int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}\,dx+\int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}\,dx.$$ Let $$J_1=\int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}\,dx.$$ Use the substitution $$x=\frac{t}{1-t}, \qquad t=\frac{x}{1+x}, \qquad dx=\frac{dt}{(1-t)^2}.$$ When $x=0$, $t=0$; when $x=1$, $t=\frac12$. Also, $$x^{m-1}=\left(\frac{t}{1-t}\right)^{m-1}, \qquad (1+x)^{m+n}=\left(\frac1{1-t}\right)^{m+n}.$$ Thus, $$J_1=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt.$$ Similarly, $$J_2=\int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}\,dx=\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt.$$ Hence, $$J=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt+\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt.$$ 3. In the second integral, put $t=1-u$. Then when $t=0$, $u=1$, and when $t=\frac12$, $u=\frac12$. So, $$\int_0^{1/2} t^{n-1}(1-t)^{m-1}\,dt =\int_{1/2}^1 u^{m-1}(1-u)^{n-1}\,du.$$ Therefore, $$J=\int_0^{1/2} t^{m-1}(1-t)^{n-1}\,dt+\int_{1/2}^1 t^{m-1}(1-t)^{n-1}\,dt.$$ So, $$J=\int_0^1 t^{m-1}(1-t)^{n-1}\,dt=I_{m,n}.$$ Thus, $$\alpha=1.$$ 4. Final answer: $$\boxed{1}$$ 5. Comparison with stored correct answer: Stored correct answer = $1$. Our derived answer matches it.