Q106 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
$\int_0^1 x(1-x)^{99} d x=$
- A$\dfrac{1}{10010}$
- Correct answerB$\dfrac{1}{10100}$
- C$-\dfrac{1}{10100}$
- D$\dfrac{1}{1010}$
Solution
Let $I = \int_{0}^{1} x(1-x)^{99} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we have:
$I = \int_{0}^{1} (1-x)(1-(1-x))^{99} dx$
$I = \int_{0}^{1} (1-x)x^{99} dx$
$I = \int_{0}^{1} (x^{99} - x^{100}) dx$
Evaluating the integral:
$I = \left[ \dfrac{x^{100}}{100} - \dfrac{x^{101}}{101} \right]_{0}^{1}$
$I = \dfrac{1}{100} - \dfrac{1}{101}$
$I = \dfrac{101 - 100}{100 \times 101}$
$I = \dfrac{1}{10100}$
Answer: $\dfrac{1}{10100}$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we have:
$I = \int_{0}^{1} (1-x)(1-(1-x))^{99} dx$
$I = \int_{0}^{1} (1-x)x^{99} dx$
$I = \int_{0}^{1} (x^{99} - x^{100}) dx$
Evaluating the integral:
$I = \left[ \dfrac{x^{100}}{100} - \dfrac{x^{101}}{101} \right]_{0}^{1}$
$I = \dfrac{1}{100} - \dfrac{1}{101}$
$I = \dfrac{101 - 100}{100 \times 101}$
$I = \dfrac{1}{10100}$
Answer: $\dfrac{1}{10100}$
Hence, the correct answer is option: 'B'.