Question

If \[ \int_{0}^{1}\left(\sum_{r=1}^{2013}\frac{x}{x^{2}+r^{2}}\right)\left(\prod_{r=1}^{2013}(x^{2}+r^{2})\right)dx =\frac{1}{2}\left[\left(\prod_{r=1}^{2013}(1+r^{2})\right)-K\right], \] then \(K\) is:

• A. $\frac{2013(2014)(4027)}{6}$
• B. $(2013)^{2013}$
• C. $(2013)!$
• D. $((2013)!)^{2}$

Hint:

Don't let giant numbers like 2013 make you nervous! They are just placeholders indicating that you should look for a structural derivative pattern. Identifying the general form $2x/(x^2+r^2)$ allows you to solve the entire problem using basic calculus rules.

Answer 1

The Correct Option is D

Concept: This problem can be solved by recognizing that the integrand expression is the exact derivative of the product function term. Let us use the standard derivative product rule for a collection of functions: $$\frac{d}{dx}\left[ f_1(x)f_2(x)\dots f_n(x) \right] = \left( \sum_{i=1}^n \frac{f_i'(x)}{f_i(x)} \right) \cdot \left( \prod_{i=1}^n f_i(x) \right)$$ Step 1: Identify the total derivative structure.
Let our product function be defined as $y = \prod_{r=1}^{2013}(x^2 + r^2)$. Let us differentiate this expression with respect to $x$ using the logarithmic product rule: $$\ln y = \sum_{r=1}^{2013} \ln(x^2 + r^2)$$ Differentiate both sides with respect to $x$: $$\frac{1}{y}\frac{dy}{dx} = \sum_{r=1}^{2013} \frac{2x}{x^2 + r^2} \quad \Rightarrow \quad \frac{dy}{dx} = 2 \cdot \left( \sum_{r=1}^{2013} \frac{x}{x^2 + r^2} \right) \cdot y$$ Substitute our original definition of $y$ back into the derivative equation: $$\frac{d}{dx}\left[ \prod_{r=1}^{2013}(x^2 + r^2) \right] = 2 \cdot \left(\sum_{r=1}^{2013}\frac{x}{x^{2}+r^{2}}\right)\left(\prod_{r=1}^{2013}(x^{2}+r^{2})\right)$$

Step 2: Substitute the derivative back into the integral.
Notice that the integrand expression matches our derived derivative formula exactly, except for a factor of 2. We can rewrite the integral as: $$I = \int_{0}^{1} \frac{1}{2} \cdot \frac{d}{dx}\left[ \prod_{r=1}^{2013}(x^2 + r^2) \right] dx$$ By the Fundamental Theorem of Calculus, the integral of a derivative simplifies directly to its boundary evaluation values: $$I = \frac{1}{2} \left[ \prod_{r=1}^{2013}(x^2 + r^2) \right]_{0}^{1}$$

Step 3: Evaluate the expression at the integration boundaries.
Substitute the upper and lower limits into the expression: $$I = \frac{1}{2} \left[ \left(\prod_{r=1}^{2013}(1^2 + r^2)\right) - \left(\prod_{r=1}^{2013}(0^2 + r^2)\right) \right]$$ $$I = \frac{1}{2} \left[ \left(\prod_{r=1}^{2013}(1 + r^2)\right) - \left(\prod_{r=1}^{2013} r^2 \right) \right] \quad \cdots (1)$$

Step 4: Simplify the lower bound product to find $K$.
Let us expand the product term at the lower bound: $$\prod_{r=1}^{2013} r^2 = 1^2 \times 2^2 \times 3^2 \times \dots \times 2013^2$$ We can group the squared terms together as a single factorial expression: $$= (1 \times 2 \times 3 \times \dots \times 2013)^2 = ((2013)!)^2$$ Now substitute this back into our integrated expression (1): $$I = \frac{1}{2} \left[ \left(\prod_{r=1}^{2013}(1 + r^2)\right) - ((2013)!)^2 \right]$$ Comparing this directly with the given answer template form, the constant $K$ must be: $$K = ((2013)!)^2$$ This matches option (D) perfectly.