Let g(x) = int_{0}^{x} f(t) dt,where f is a c | vedclass

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(A) Given that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$.We need to find the ran

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$[\frac{1}{3}, 2]$

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(A) Given that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$.We need to find the range of $g(3) = \int_{0}^{3} f(t) dt$.We can split the integral as $g(3) = \int_{0}^{1} f(t) dt + \int_{1}^{3} f(t) dt$.For the first part,$\int_{0}^{1} \frac{1}{3} dt \leq \int_{0}^{1} f(t) dt \leq \int_{0}^{1} 1 dt$,which gives $\frac{1}{3} \leq \int_{0}^{1} f(t) dt \leq 1$.For the second part,$\int_{1}^{3} 0 dt \leq \int_{1}^{3} f(t) dt \leq \int_{1}^{3} \frac{1}{2} dt$,which gives $0 \leq \int_{1}^{3} f(t) dt \leq \frac{1}{2} 	imes (3 - 1) = 1$.Adding these two inequalities,we get $\frac{1}{3} + 0 \leq g(3) \leq 1 + 1$,which simplifies to $\frac{1}{3} \leq g(3) \leq 2$.Thus,the interval is $[\frac{1}{3}, 2]$.

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

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