Practicing Success
If $\int\limits_a^b\frac{x^n}{x^n+(16-x)^n}dx=6$ then: |
a = 4, b = 12, n ∈ R $a = 2, b = 14, n ∈ R$ a = -4, b = 20, n ∈ R a = 2, b = 8, n ∈ R |
$a = 2, b = 14, n ∈ R$ |
$I=\int\limits_a^b\frac{x^n}{x^n+(16-x)^n}dx=\int\limits_a^b\frac{(a+b-x)^n}{(a+b-x)^n+(16-a-b+x)^n}dx$ $I=\int\limits_a^b\frac{x^n}{x^n+(16-x)^n}dx=\int\limits_a^b\frac{(16-x)^n}{(16-x)^n+x^n}dx$ [only way I = 6 ∀ n ∈ R possible when a + b = 16] $⇒I=\frac{1}{2}\int\limits_a^b\frac{x^n+(16-x)^n}{x^n+(16-x)^n}dx=\frac{b-a}{2}=6$; a + b = 16 and b – a = 12 $⇒ a = 2, b = 14n ∈ R$ |