Let g(x)=∫ _0^xf(t)dt, where f is continuous function in [0,3] such that 1 3≤ f(t)≤ 1 for all t∈ [0,1] and 0≤ f(t)≤ 1 2 for all t∈ (1,3]. The largest…

$$\frac{1}{3}\leq f(t)\leq 1\forall t\in [0,1]$$31​≤f(t)≤1∀t∈[0,1]

$$0\leq f(t)\leq \frac{1}{2}\forall t\in (1,3]$$0≤f(t)≤21​∀t∈(1,3]

Now, $\displaystyle g(3)={\int }_{0}^{3}f(t)dt={\int }_{0}^{1}f(t)dt+{\int }_{1}^{3}f(t)dt$g(3)=∫03​f(t)dt=∫01​f(t)dt+∫13​f(t)dt

$$\because {\int }_{0}^{1}\frac{1}{3}dt\leq {\int }_{0}^{1}f(t)dt\leq {\int }_{0}^{1}1.dt...(1)$$∵∫01​31​dt≤∫01​f(t)dt≤∫01​1.dt...(1)

and $\displaystyle {\int }_{1}^{3}0dt\leq {\int }_{1}^{3}f(1)dt\leq {\int }_{1}^{3}\frac{1}{2}dt...(2)$∫13​0dt≤∫13​f(1)dt≤∫13​21​dt...(2)

Adding, we get

$$\frac{1}{3}+0\leq g(3)\leq 1+\frac{1}{2}(3-1)$$31​+0≤g(3)≤1+21​(3−1)

$$\frac{1}{3}\leq g(3)\leq 2$$31​≤g(3)≤2