If `int_(a)^(b) (x^(n))/(x^(n)+(16-x)^(n))dx=6`, then

Options

A. `a=4,b=12,n in R`
B. `a=2,b=14,n in R`
C. `a=-4,b=20,n in R`
D. `a=2,b=8, n in R`

Answer

To solve the given integral equation

\[ \int_{a}^{b} \frac{x^n}{x^n + (16 - x)^n} \, dx = 6, \]

we can use the property of definite integrals that states:

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

Here, we will define

\[ f(x) = \frac{x^n}{x^n + (16 - x)^n}. \]

Now, let's compute \( f(16 - x) \):

\[ f(16 - x) = \frac{(16 - x)^n}{(16 - x)^n + x^n}. \]

Notice that:

\[ f(x) + f(16 - x) = \frac{x^n}{x^n + (16 - x)^n} + \frac{(16 - x)^n}{(16 - x)^n + x^n} = 1. \]

Thus, we can write:

\[ \int_{a}^{b} f(x) \, dx + \int_{a}^{b} f(16 - x) \, dx = \int_{a}^{b} 1 \, dx = b - a. \]

Since both integrals are equal, we can denote:

\[ I = \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(16 - x) \, dx. \]

Therefore, we have:

\[ I + I = b - a \quad \Rightarrow \quad 2I = b - a \quad \Rightarrow \quad I = \frac{b - a}{2}. \]

Given that \( I = 6 \), we can substitute:

\[ 6 = \frac{b - a}{2} \quad \Rightarrow \quad b - a = 12. \]

Now, we can express \( b \) in terms of \( a \):

\[ b = a + 12. \]

Next, we need to find the values of \( a \) and \( b \). Since \( a \) and \( b \) are not specified in the problem, they can take any values as long as their difference is 12. For instance, we can choose \( a = 0 \) and \( b = 12 \).

Thus, we can conclude:

\[ a = 0, \quad b = 12. \]

Now, we need to determine the value of \( n \). The integral \( \int_{0}^{12} \frac{x^n}{x^n + (16 - x)^n} \, dx = 6 \) holds for any positive integer \( n \). However, for the sake of simplicity, we can choose \( n = 1 \) as a common case.

### Final Answer:
- \( a = 0 \)
- \( b = 12 \)
- \( n = 1 \)