Question

$\int_{0}^{5}\frac{(x+8)^{2026}}{(x+8)^{2026}+(13-x)^{2026}}dx$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{3}{4}$
• C. 5
• D. $\frac{5}{4}$
• E. $\frac{5}{2}$

Hint:

Math Tip: Whenever you encounter a definite integral structured as $\int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx$, it will always symmetrically fold into itself. The answer is universally exactly half the length of the integration interval: $\frac{b - a}{2}$. Here, $\frac{5 - 0}{2} = \frac{5}{2}$.

Answer 1

Answer: (E)

Concept:
Calculus - Definite Integration Properties.
King's Rule: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.
Step 1: Set up the initial equation.
Let the given integral be equal to $I$: $$ I = \int_{0}^{5} \frac{(x+8)^{2026}}{(x+8)^{2026} + (13-x)^{2026}} dx \quad \text{--- (Equation 1)} $$
Step 2: Apply King's Rule.
The limits are $a = 0$ and $b = 5$. The sum of the limits is $a + b = 5$. Replace every $x$ in the integral with $(5 - x)$: $$ I = \int_{0}^{5} \frac{((5-x)+8)^{2026}}{((5-x)+8)^{2026} + (13-(5-x))^{2026}} dx $$
Step 3: Simplify the replaced terms.

• $(5 - x) + 8 = 13 - x$
• $13 - (5 - x) = 13 - 5 + x = 8 + x$

Substitute these simplified terms back into the integral: $$ I = \int_{0}^{5} \frac{(13-x)^{2026}}{(13-x)^{2026} + (x+8)^{2026}} dx \quad \text{--- (Equation 2)} $$
Step 4: Add Equation 1 and Equation 2.
Adding the left sides gives $2I$. Since the denominators of both integrals are identical, we can directly add their numerators: $$ 2I = \int_{0}^{5} \frac{(x+8)^{2026} + (13-x)^{2026}}{(x+8)^{2026} + (13-x)^{2026}} dx $$
Step 5: Evaluate the simplified integral.
The numerator and denominator perfectly cancel out, leaving exactly $1$: $$ 2I = \int_{0}^{5} 1 \,dx $$ $$ 2I = [x]_{0}^{5} $$ $$ 2I = 5 - 0 = 5 $$
Step 6: Solve for I.
Divide both sides by 2 to isolate $I$: $$ I = \frac{5}{2} $$