Suppose g(x) = int_0^x f(t) dt,where f is suc | vedclass

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(A) We are given $g(2) = \int_0^2 f(t) dt = \int_0^1 f(t) dt + \int_1^2 f(t) dt$. For $t \in [0, 1]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating

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$\frac{1}{2} \le g(2) \le \frac{3}{2}$

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(A) We are given $g(2) = \int_0^2 f(t) dt = \int_0^1 f(t) dt + \int_1^2 f(t) dt$. For $t \in [0, 1]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating this,we get $\int_0^1 0 dt \le \int_0^1 f(t) dt \le \int_0^1 \frac{1}{2} dt$,which simplifies to $0 \le \int_0^1 f(t) dt \le \frac{1}{2}$. For $t \in [1, 2]$,we have $\frac{1}{2} \le f(t) \le 1$. Integrating this,we get $\int_1^2 \frac{1}{2} dt \le \int_1^2 f(t) dt \le \int_1^2 1 dt$,which simplifies to $\frac{1}{2} \le \int_1^2 f(t) dt \le 1$. Adding these two inequalities,we get $0 + \frac{1}{2} \le \int_0^1 f(t) dt + \int_1^2 f(t) dt \le \frac{1}{2} + 1$. Therefore,$\frac{1}{2} \le g(2) \le \frac{3}{2}$.

Suppose $g(x) = \int_0^x f(t) dt$,where $f$ is such that for $t \in [0, 1]$,$0 \le f(t) \le \frac{1}{2}$ and for $t \in [1, 2]$,$\frac{1}{2} \le f(t) \le 1$. Find the range of $g(2)$.

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