Question:
Evaluate the definite integral: $\int_{0}^{2026} \frac{x^5}{x^5 + (2026 - x)^5} \, dx$
Evaluate the definite integral: $\int_{0}^{2026} \frac{x^5}{x^5 + (2026 - x)^5} \, dx$
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Keep this shortcut memorized for competitive exams! Whenever you encounter a definite integral structured as \(\int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx\), the function expression will *always* simplify to unity upon addition. The value of the integral is simply half the total interval length: \(\frac{b - a}{2}\). For this problem: \(\frac{2026 - 0}{2} = 1013\) instantly!
Updated On: May 26, 2026
- \( 2026 \)
- \( 1013 \)
- \( 506.5 \)
- \( 0 \)
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The Correct Option is B
Solution and Explanation
Concept:
To solve definite integrals involving complex, symmetric fractional functions where standard algebraic integration looks impossible, we invoke King's Property of Definite Integrals:
\[
\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx
\]
Step 1: Applying King's Property to set up a twin integral equation.
Let our target definite integral equation be marked as \(I\): \[ I = \int_{0}^{2026} \frac{x^5}{x^5 + (2026 - x)^5} \, dx \quad \cdots \text{(Equation 1)} \] Applying King's Property, we replace every instance of \(x\) across the function boundary with \((0 + 2026 - x) = (2026 - x)\): \[ I = \int_{0}^{2026} \frac{(2026 - x)^5}{(2026 - x)^5 + (2026 - (2026 - x))^5} \, dx \] Simplifying the interior bracket terms yields: \[ I = \int_{0}^{2026} \frac{(2026 - x)^5}{(2026 - x)^5 + x^5} \, dx \quad \cdots \text{(Equation 2)} \]
Step 2: Adding both integral representations together.
Since Equation 1 and Equation 2 have identical upper and lower integration limits, we can sum their integrands directly: \[ 2I = \int_{0}^{2026} \left[ \frac{x^5}{x^5 + (2026 - x)^5} + \frac{(2026 - x)^5}{x^5 + (2026 - x)^5} \right] dx \] Notice that the numerators combine to perfectly match the common denominator: \[ 2I = \int_{0}^{2026} \frac{x^5 + (2026 - x)^5}{x^5 + (2026 - x)^5} \, dx \] \[ 2I = \int_{0}^{2026} 1 \cdot dx \]
Step 3: Integrating and isolating $I$.
Evaluating the simple integration constant across the limits: \[ 2I = \Big[ x \Big]_{0}^{2026} = 2026 - 0 = 2026 \] \[ I = \frac{2026}{2} = 1013 \]
Step 1: Applying King's Property to set up a twin integral equation.
Let our target definite integral equation be marked as \(I\): \[ I = \int_{0}^{2026} \frac{x^5}{x^5 + (2026 - x)^5} \, dx \quad \cdots \text{(Equation 1)} \] Applying King's Property, we replace every instance of \(x\) across the function boundary with \((0 + 2026 - x) = (2026 - x)\): \[ I = \int_{0}^{2026} \frac{(2026 - x)^5}{(2026 - x)^5 + (2026 - (2026 - x))^5} \, dx \] Simplifying the interior bracket terms yields: \[ I = \int_{0}^{2026} \frac{(2026 - x)^5}{(2026 - x)^5 + x^5} \, dx \quad \cdots \text{(Equation 2)} \]
Step 2: Adding both integral representations together.
Since Equation 1 and Equation 2 have identical upper and lower integration limits, we can sum their integrands directly: \[ 2I = \int_{0}^{2026} \left[ \frac{x^5}{x^5 + (2026 - x)^5} + \frac{(2026 - x)^5}{x^5 + (2026 - x)^5} \right] dx \] Notice that the numerators combine to perfectly match the common denominator: \[ 2I = \int_{0}^{2026} \frac{x^5 + (2026 - x)^5}{x^5 + (2026 - x)^5} \, dx \] \[ 2I = \int_{0}^{2026} 1 \cdot dx \]
Step 3: Integrating and isolating $I$.
Evaluating the simple integration constant across the limits: \[ 2I = \Big[ x \Big]_{0}^{2026} = 2026 - 0 = 2026 \] \[ I = \frac{2026}{2} = 1013 \]
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