If $\int_0^n[x] \mathrm{dx}=66$, then $n=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^n[x] d x=\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\ldots \ldots+ \ & \int_{n-1}^n(n-1) d x \ & =1+2+3+\ldots .+(n-1)=\frac{(n-1) n}{2}=66 \ & \Rightarrow n(n-1)=132 \Rightarrow n=12 \end{aligned}$

If \(\int_0^n[x] \mathrm{dx}=66\), then \(n=\)

Solution