Question

integral of (1 - x⁵⁰)¹⁰⁰ from 0 to 1 divided by integral of (1 - x⁵⁰)¹⁰¹ from 0 to 1

Answer 1

$$\begin{gathered} \frac{{\int }_0^1{\left(1-x^{50}\right)}^{100}dx}{{\int }_0^1{\left(1-x^{50}\right)}^{101}dx}=\frac{{\int }_0^{1}1\times {\left(1-x^{50}\right)}^{100}dx}{{\int }_0^{1}1\times {\left(1-x^{50}\right)}^{101}dx} \\ =\frac{{\left(1-x^{50}\right)}^{100}{\int }_0^{1}1dx-{\int }_0^1\left({\displaystyle \frac{d}{dx}}{\left(1-x^{50}\right)}^{100}\left({\int }_{}^{}1dx\right)dx\right)}{{\left(1-x^{50}\right)}^{101}{\int }_0^{1}1dx-{\int }_0^1\left({\displaystyle \frac{d}{dx}}{\left(1-x^{50}\right)}^{101}\left({\int }_{}^{}1dx\right)dx\right)} \\ =\frac{{\left[{\left(1-x^{50}\right)}^{100}x\right]}_0^1-{\int }_0^1\left(100{\left(1-x^{50}\right)}^{99}\left(-50x^{49}\right)\left(x\right)\right)dx}{{\left[{\left(1-x^{50}\right)}^{101}x\right]}_0^1-{\int }_0^1\left(101{\left(1-x^{50}\right)}^{100}\left(-50x^{49}\right)\left(x\right)\right)dx} \end{gathered}$$
$$\begin{gathered} =\frac{0-{\int }_0^1\left(100{\left(1-x^{50}\right)}^{99}(-50x^{50})\right)dx}{0-{\int }_0^{1}101{\left(1-x^{50}\right)}^{100}(-50x^{50})dx} \\ =\frac{100{\int }_0^1{\left(1-x^{50}\right)}^{99}\left(x^{50}\right)dx}{101{\int }_0^1{\left(1-x^{50}\right)}^{100}\left(x^{50}\right)dx} \\ Thiswillgoonas,I={\displaystyle \frac{100}{101}}\times {\displaystyle \frac{99}{100}}\times {\displaystyle \frac{98}{99}}\times {\displaystyle \frac{97}{98}}\times {\displaystyle \frac{96}{97}}\times {\displaystyle \frac{95}{96}}\times \cdots \times \frac{{\int }_0^1{\left(1-x^{50}\right)}^0\left(x^{50}\right)dx}{{\int }_0^1{\left(1-x^{50}\right)}^1\left(x^{50}\right)dx} \\ I={\displaystyle \frac{100}{101}}\times {\displaystyle \frac{99}{100}}\times {\displaystyle \frac{98}{99}}\times {\displaystyle \frac{97}{98}}\times {\displaystyle \frac{96}{97}}\times {\displaystyle \frac{95}{96}}\times \cdots \times \frac{{\int }_0^1\left(x^{50}\right)dx}{{\int }_0^1{\left(1-x^{50}\right)}^1\left(x^{50}\right)dx} \\ I=\frac{100}{101}\times \frac{99}{100}\times \frac{98}{99}\times \frac{97}{98}\times \frac{96}{97}\times \frac{95}{96}\times \cdots \times \frac{{\left[\frac{x^{51}}{51}\right]}_0^1}{{\left[\frac{x^{51}}{51}\right]}_0^1-{\left[{\displaystyle \frac{x^{101}}{101}}\right]}_0^1} \end{gathered}$$

$$\begin{gathered} I=\frac{100}{101}\times \frac{99}{100}\times \frac{98}{99}\times \frac{97}{98}\times \frac{96}{97}\times \frac{95}{96}\times \cdots \times \frac{{\left[\frac{x^{51}}{51}\right]}_0^1}{{\left[\frac{x^{51}}{51}\right]}_0^1-{\left[{\displaystyle \frac{x^{101}}{101}}\right]}_0^1} \\ I=\frac{100}{101}\times \frac{99}{100}\times \frac{98}{99}\times \frac{97}{98}\times \frac{96}{97}\times \frac{95}{96}\times \cdots \times {\displaystyle \frac{1}{2}}\times \frac{{\displaystyle \frac{1}{51}}}{{\displaystyle \frac{1}{51}}-{\displaystyle \frac{1}{101}}} \end{gathered}$$

$$\begin{gathered} I=\frac{100}{101}\times \frac{99}{100}\times \frac{98}{99}\times \frac{97}{98}\times \frac{96}{97}\times \frac{95}{96}\times \cdots \times {\displaystyle \frac{1}{2}}\times \frac{{\displaystyle \frac{1}{51}}}{{\displaystyle \frac{50}{51\times 101}}} \\ I=\frac{1}{101}\times \frac{101}{50}=\frac{1}{50} \end{gathered}$$