The value of $\int_0^{2^n}[\sqrt{x}]dx$, ([*] denotes the greatest integer function), $n ∈ N$, is
Answer & explanation
Correct answer: option 2
$\int_0^{n^2}[\sqrt{x}]dx=\int_0^{1^2}[\sqrt{x}]dx+\int_{1^2}^{2^2}[\sqrt{x}]dx+\int_{2^2}^{3^2}[\sqrt{x}]dx+...+\int_{(n-1)^2}^{n^2}[\sqrt{x}]dx$
$= 0 + 1. (2^2 – 1^2) + 2. (3^2 – 2^2) + ....$
$+....+ (n - 1) [n^2 – (n – 1)^2]$
$–1^2 – 2^2 – 3^3–.....–(n – 1)^2 + (n – 1)n^2$
$= – (1^2 + 2^2 + 3^2+.....+(n – 1)^2 + (n – 1)n^2$
$=-\frac{(n-1)n(2n-1)}{6}+ (n – 1) n^2$
$=\frac{1}{6}n (n – 1) (4n + 1)$