Let f:(2, infty) rightarrow mathbb{N} be defi | vedclass

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(B) Given $f(x) = 	ext{largest prime factor of } [x]$.We need to evaluate $I = \int_{2}^{8} f(x) \, dx$.Since $[x]$ is constant on intervals $[n, n+1

Vedclass · Accepted Answer

$22$

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(B) Given $f(x) = 	ext{largest prime factor of } [x]$.We need to evaluate $I = \int_{2}^{8} f(x) \, dx$.Since $[x]$ is constant on intervals $[n, n+1)$,we split the integral:$I = \int_{2}^{3} f(x) \, dx + \int_{3}^{4} f(x) \, dx + \int_{4}^{5} f(x) \, dx + \int_{5}^{6} f(x) \, dx + \int_{6}^{7} f(x) \, dx + \int_{7}^{8} f(x) \, dx$.For $x \in [2, 3)$,$[x] = 2$,largest prime factor is $2$.For $x \in [3, 4)$,$[x] = 3$,largest prime factor is $3$.For $x \in [4, 5)$,$[x] = 4$,largest prime factor is $2$.For $x \in [5, 6)$,$[x] = 5$,largest prime factor is $5$.For $x \in [6, 7)$,$[x] = 6$,largest prime factor is $3$.For $x \in [7, 8)$,$[x] = 7$,largest prime factor is $7$.Thus,$I = \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{5} 2 \, dx + \int_{5}^{6} 5 \, dx + \int_{6}^{7} 3 \, dx + \int_{7}^{8} 7 \, dx$.$I = 2(1) + 3(1) + 2(1) + 5(1) + 3(1) + 7(1) = 2 + 3 + 2 + 5 + 3 + 7 = 22$.

Let $f:(2, \infty) \rightarrow \mathbb{N}$ be defined by $f(x) =$ the largest prime factor of $[x]$. Then,$\int_{2}^{8} f(x) \, dx$ is equal to

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