दिया है, \(\mathrm{I}=\int_{0}^{1} x(1-x)^{99} d x\)
जैसा कि हम जानते हैं,
\(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
अत: \(\mathrm{I}=\int_{0}^{1}(1-x)\{1-(1-x)\}^{99} d \)
\(x=\int_{0}^{1}(1-x)(1-1+x)^{99} d x\)
\(=\int_{0}^{1}(1-x)(x)^{99} d x=\int_{0}^{1}\left(x^{99}-x^{100}\right) d x\)
\(=\int_{0}^{1}(x)^{99} d x-\int_{0}^{1}(x)^{99} d x\)
\(=\left[\frac{x^{100}}{100}\right]_{0}^{1}\)
\(=\left[\frac{x^{101}}{101}\right]_{0}^{1}\)
\(=\left[\frac{1^{100}-0}{100}\right]-\left[\frac{1^{101}-0}{101}\right]\)
\(=\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{10100}\)
