Let g(x) = [_0^x f(t)dt ], where f is continuous function in [ 0, 3 ] such that [1 3] [ ] …

Question

Let g(x) = $$\int_0^x {f(t)dt} $$, where f is continuous function in [ 0, 3 ] such that $${1 \over 3}$$ $$ \le $$ f(t) $$ \le $$ 1 for all t$$\in$$ [0, 1] and 0 $$ \le $$ f(t) $$ \le $$ $${1 \over 2}$$ for all t$$\in$$ (1, 3]. The largest possible interval in which g(3) lies is :

Accepted Answer

Given, $g(x)=\int_0^x f(t) d t$

$	herefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t$

$\Rightarrow \int_0^1 \frac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \frac{1}{2} d t$

$\Rightarrow \frac{1}{3} \leq g(3) \leq 1+1$

$\Rightarrow \frac{1}{3} \leq g(3) \leq 2$

Solutions · 1