Integral Calculus
Definite Integral
MMTS_Full_Test_13
Grade 12
Question:
$\int_{-5}^{5}\dfrac{x^{10}}{x^{10}+(5-x)^{10}}\cdot\dfrac{dx}{x^2+16}$
$\dfrac{1}{4}\tan^{-1}\dfrac{5}{4}$
$\dfrac{1}{2}\tan^{-1}\dfrac{5}{4}$
$2\tan^{-1}\dfrac{5}{4}$
$\tan^{-1}\dfrac{5}{4}$
Step-by-Step Solution
Key Concept: Use $f(x)+f(-x)$ symmetry: $\frac{x^{10}}{x^{10}+(5-x)^{10}}+\frac{(-x)^{10}}{(-x)^{10}+(5-(-x))^{10}}=?$
Let $I=\int_{-5}^5\frac{x^{10}}{x^{10}+(5-x)^{10}}\cdot\frac{dx}{x^2+16}$. Sub $x\to -x$: $I=\int_{-5}^5\frac{x^{10}}{x^{10}+(5+x)^{10}}\cdot\frac{dx}{x^2+16}$. Adding: $2I=\int_{-5}^5\frac{dx}{x^2+16}=\frac{1}{4}\tan^{-1}(x/4)|_{-5}^5=\frac{1}{2}\tan^{-1}(5/4)$. $I=\frac{1}{4}\tan^{-1}\frac{5}{4}$.
Correct Answer: 4