If I_1=_0^1(1-x^50)^100dx and I_2=_0^1(1-x^50)^101dx such that I_2= I_1 then equals to :

Question

If $I_1=\displaystyle\int_0^1(1-x^{50})^{100}dx$ and $I_2=\displaystyle\int_0^1(1-x^{50})^{101}dx$ such that $I_2=\alpha I_1$ then $\alpha$ equals to :

Accepted Answer

$$I_2=\int_0^1(1-x^{50})(1-x^{50})^{100}dx=I_1-\int_0^1x\cdot x^{49}(1-x^{50})^{100}dx$$

Integrating the second term by parts (with $u=(1-x^{50})^{100}$):
$$\int_0^1x^{49}(1-x^{50})^{100}dx=\left[\frac{x^{50}}{5050}(1-x^{50})^{100}\right]_0^1... $$

Using the by-parts route instead: $I_2=I_1+\left[\dfrac{x^{50}}{5050}(1-x^{50})^{101}\right]_0^1-\dfrac1{5050}\displaystyle\int_0^1(1-x^{50})^{101}dx=I_1+0-\dfrac{I_2}{5050}$

$$\Rightarrow\frac{5051}{5050}I_2=I_1\Rightarrow I_2=\frac{5050}{5051}I_1\Rightarrow\alpha=\frac{5050}{5051}$$

Solutions · 1