`int_0^10 (x^10)/((10 - x)^10 + x^10)` dx is equal to

Options

A. 10
B. 5
C. 2
D. `1/2`

Answer

To solve the integral

\[ I = \int_0^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} \, dx, \]

we can utilize a property of definite integrals.

### Step 1: Use the substitution property of integrals

We can use the substitution \( x = 10 - t \). Then, \( dx = -dt \). When \( x = 0 \), \( t = 10 \) and when \( x = 10 \), \( t = 0 \). Therefore, we can rewrite the integral as:

\[ I = \int_{10}^{0} \frac{(10 - t)^{10}}{(10 - (10 - t))^{10} + (10 - t)^{10}} (-dt) = \int_0^{10} \frac{(10 - t)^{10}}{t^{10} + (10 - t)^{10}} \, dt. \]

### Step 2: Rewrite the integral

Now, we can denote this new integral as \( I \):

\[ I = \int_0^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} \, dx. \]

### Step 3: Add the two expressions for \( I \)

Now we have two expressions for \( I \):

1. \( I = \int_0^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} \, dx \)
2. \( I = \int_0^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} \, dx \)

Adding these two equations gives:

\[ 2I = \int_0^{10} \left( \frac{x^{10}}{(10 - x)^{10} + x^{10}} + \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} \right) \, dx. \]

### Step 4: Simplify the expression

Notice that the denominators are the same, so we can combine the fractions:

\[ 2I = \int_0^{10} \frac{x^{10} + (10 - x)^{10}}{(10 - x)^{10} + x^{10}} \, dx = \int_0^{10} 1 \, dx. \]

### Step 5: Evaluate the integral

Now, we can evaluate the integral:

\[ 2I = \int_0^{10} 1 \, dx = [x]_0^{10} = 10 - 0 = 10. \]

### Step 6: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = \frac{10}{2} = 5. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{5}. \]