Step 1: Understanding the Concept:
This is a definite integral that can be solved using properties of definite integrals or by recognizing it as a form of the Beta function. The property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ is particularly useful here.
Step 2: Key Formula or Approach:
Method 1: Property of Definite Integrals
Let $I = \int_0^1 x(1-x)^9 \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$, with $a=1$:
\[ I = \int_0^1 (1-x)(1-(1-x))^9 \,dx \]
Method 2: Beta Function
The Beta function is defined as $B(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} \,dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. For integer values, $\Gamma(k)=(k-1)!$.
Step 3: Detailed Explanation (Method 1):
Let $I = \int_0^1 x(1-x)^9 \,dx$.
Apply the property with $a=1$:
\[ I = \int_0^1 (1-x)(1-(1-x))^9 \,dx \]
\[ I = \int_0^1 (1-x)(x)^9 \,dx \]
\[ I = \int_0^1 (x^9 - x^{10}) \,dx \]
Now, integrate term by term:
\[ I = \left[ \frac{x^{10}}{10} - \frac{x^{11}}{11} \right]_0^1 \]
Evaluate at the limits:
\[ I = \left( \frac{1^{10}}{10} - \frac{1^{11}}{11} \right) - \left( \frac{0^{10}}{10} - \frac{0^{11}}{11} \right) \]
\[ I = \frac{1}{10} - \frac{1}{11} = \frac{11 - 10}{110} = \frac{1}{110} \]
Step 3: Detailed Explanation (Method 2):
The integral is $I = \int_0^1 x^1 (1-x)^9 \,dx$.
This matches the Beta function form $\int_0^1 x^{m-1}(1-x)^{n-1} \,dx$ with $m-1=1 \implies m=2$ and $n-1=9 \implies n=10$.
\[ I = B(2, 10) = \frac{\Gamma(2)\Gamma(10)}{\Gamma(2+10)} = \frac{\Gamma(2)\Gamma(10)}{\Gamma(12)} \]
Using $\Gamma(k) = (k-1)!$:
\[ I = \frac{1! \cdot 9!}{11!} = \frac{1 \cdot 9!}{11 \cdot 10 \cdot 9!} = \frac{1}{11 \cdot 10} = \frac{1}{110} \]
Step 4: Final Answer:
Both methods give the value $\frac{1}{110}$. Therefore, option (A) is correct.