If int_1^n [x] dx = 120,then n = | vedclass

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(B) Given the integral $\int_1^n [x] dx = 120$. We can split the integral into intervals of unit length: $\int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 d

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$16$

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(B) Given the integral $\int_1^n [x] dx = 120$. We can split the integral into intervals of unit length: $\int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 dx + \dots + \int_{n-1}^n (n-1) dx = 120$. This simplifies to the sum of the first $(n-1)$ natural numbers: $1 + 2 + 3 + \dots + (n-1) = 120$. Using the formula for the sum of the first $k$ natural numbers,$\frac{k(k+1)}{2}$,where $k = n-1$: $\frac{(n-1)n}{2} = 120$. $(n-1)n = 240$. $n^2 - n - 240 = 0$. $(n - 16)(n + 15) = 0$. Since $n$ must be positive,$n = 16$.

If $\int_1^n [x] dx = 120$,then $n = $

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