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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$22$

Vedclass · Answer

(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

Similar Questions

$\int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$

$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

$\int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx =$

$\int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx =$

$\int_0^{\pi /2} {\sqrt {\cos \theta } {{\sin }^3}\theta } \,d\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module