`int_(0)^(11)((11-x)^(2))/(x^(2)+(11-x)^(2))dx=`

Options

A. 2
B. `11/2`
C. `4`
D. `1/2`

Answer

To solve the integral

\[ I = \int_{0}^{11} \frac{(11-x)^{2}}{x^{2} + (11-x)^{2}} \, dx, \]

we can use a property of definite integrals. The property states that if we have an integral from \(a\) to \(b\) of a function \(f(x)\), we can express it as:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In this case, \(a = 0\) and \(b = 11\). Therefore, we can rewrite the integral as follows:

1. **Substituting \(x\) with \(11 - x\)**:

\[ I = \int_{0}^{11} \frac{(11 - (11 - x))^{2}}{(11 - x)^{2} + (11 - (11 - x))^{2}} \, dx. \]

Simplifying the expression inside the integral gives:

\[ I = \int_{0}^{11} \frac{x^{2}}{(11-x)^{2} + x^{2}} \, dx. \]

2. **Now we have two expressions for \(I\)**:

\[ I = \int_{0}^{11} \frac{(11-x)^{2}}{x^{2} + (11-x)^{2}} \, dx, \]
\[ I = \int_{0}^{11} \frac{x^{2}}{(11-x)^{2} + x^{2}} \, dx. \]

3. **Adding these two integrals**:

\[ 2I = \int_{0}^{11} \left( \frac{(11-x)^{2}}{x^{2} + (11-x)^{2}} + \frac{x^{2}}{(11-x)^{2} + x^{2}} \right) \, dx. \]

Since the denominators are the same, we can combine the fractions:

\[ 2I = \int_{0}^{11} \frac{(11-x)^{2} + x^{2}}{x^{2} + (11-x)^{2}} \, dx. \]

The numerator simplifies to:

\[ (11-x)^{2} + x^{2} = 121 - 22x + 2x^{2}. \]

4. **The integral simplifies to**:

\[ 2I = \int_{0}^{11} 1 \, dx = 11. \]

5. **Solving for \(I\)**:

\[ 2I = 11 \implies I = \frac{11}{2}. \]

Thus, the value of the integral is

\[ \boxed{\frac{11}{2}}. \]