Let n geq 2 be a natural number and f:[0,1] r | vedclass

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(B) The area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is given by the integral $\int_{0}^{1} |f(x)| dx$. Based on the provided

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$8$

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(B) The area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is given by the integral $\int_{0}^{1} |f(x)| dx$. Based on the provided graph,the area consists of three triangular regions $I, II, III$ and a trapezoidal region. Area of region $I$ (triangle with base $\frac{1}{2n}$ and height $n$): $\frac{1}{2} 	imes \frac{1}{2n} 	imes n = \frac{1}{4}$.Area of region $II$ (triangle with base $\frac{1}{2n}$ and height $n$): $\frac{1}{2} 	imes \frac{1}{2n} 	imes n = \frac{1}{4}$.Area of region $III$ (trapezoid with parallel sides $n$ and $0$ and height $1-\frac{1}{n}$): $\frac{1}{2} 	imes (n+0) 	imes (1-\frac{1}{n}) = \frac{n}{2} 	imes \frac{n-1}{n} = \frac{n-1}{2}$.Total Area = $\frac{1}{4} + \frac{1}{4} + \frac{n-1}{2} = 4$.$\frac{1}{2} + \frac{n-1}{2} = 4 \implies \frac{n}{2} = 4 \implies n = 8$.The maximum value of $f(x)$ is $n = 8$.

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