If `A= int _(0)^(1) prod_(r=1) ^(2014)(r-x) dx and B =int _(0)^(1) prod _(r =0)^(2013) (r+x) dx, ` then:

Options

A. `A=2B`
B. `2A=B`
C. `A+B=0`
D. `A=B`

Answer

To solve the problem, we need to evaluate the integrals \( A \) and \( B \) defined as follows:

\[ A = \int_0^1 \prod_{r=1}^{2014} (r - x) \, dx \]

\[ B = \int_0^1 \prod_{r=0}^{2013} (r + x) \, dx \]

### Step 1: Evaluate \( A \)

The expression for \( A \) can be rewritten as:

\[ A = \int_0^1 (1 - x)(2 - x)(3 - x) \cdots (2014 - x) \, dx \]

This is a polynomial of degree 2014 in \( x \). The integral of a polynomial can be computed using the formula:

\[ \int_0^1 x^n \, dx = \frac{1}{n+1} \]

However, we need to express \( A \) in a more manageable form. The polynomial can be expanded, but it is more efficient to recognize that the integral represents the area under the curve of the polynomial from 0 to 1.

### Step 2: Evaluate \( B \)

The expression for \( B \) can be rewritten as:

\[ B = \int_0^1 (0 + x)(1 + x)(2 + x) \cdots (2013 + x) \, dx \]

This is also a polynomial, specifically of degree 2014.

### Step 3: Relate \( A \) and \( B \)

Notice that both \( A \) and \( B \) are integrals of polynomials of the same degree. We can use the symmetry of the polynomials to relate \( A \) and \( B \).

### Step 4: Use properties of definite integrals

Using the property of definite integrals, we can find that:

\[ A = (-1)^{2014} B = B \]

This is because the transformation \( x \to 1 - x \) will change the sign of the polynomial in \( A \).

### Step 5: Conclusion

Since both \( A \) and \( B \) evaluate to the same integral, we conclude that:

\[ A = B \]

### Final Answer

Thus, the relationship between \( A \) and \( B \) is:

\[ A = B \]