Does this idea generally true ?!

Question

Does $\int \limits_{a}^{b} f(t) dt \leq \int \limits_{a}^{b} g(t) dt \leq \int \limits_{a}^{b} h(t) dt$ Given that $f(x) \leq g(x) \leq h(x)$?

If the above is true, could we say that :

$f(b)-f(a) \leq \int \limits_{a}^{b} g(t) dt \leq h(b)-h(a)$ if $f'(x) \leq g(x) \leq h'(x)$ ?!

Answer 1

For the second inequality (assuming continuity and integrability where necessary and that $a\leq b$), since $f'(x)\leq g(x)\leq h'(x)$, we know that $$ \int_a^b f'(x)dx\leq \int_a^b g(x)dx\leq \int_a^bh'(x)dx. $$ Applying the fundamental theorem of calculus gives us $$ f(b)-f(a)\leq\int_a^b g(x)dx\leq h(b)-h(a) $$ as desired. So, yes, your statement is correct.

Answer 2

hint

Use the first Theorem of Calculus which states that

if

1. $f $ continuous at $[a,b] $.
2. $f $ differentiable at $(a,b) $.
3. $f'$ integrable at $[a,b] $

then $$\int_a^bf'(x)dx=f (b)-f (a) $$