Hi guys was doing the following question:
$h$ is a Riemann integrable function on $[a,b]$. Suppose that there exists a $m>0$ such that $h(x)\ge m$ for all values of $x\in[a,b]$. Prove $\int^b_ah(x)\,dx>0$.
My attempt at this proof takes into account that since the function values of h(x) $\geq$ m where m is a positive number therefore the function lies above the x axis i interpret this function as one having a local minimum at a point m. Also i consider since the graph or function is above the x axis the integral will be positive based on the value m > 0 and being the smallest value in the function.
Therefore i state the following:-
since the h(x) is Riemann Integrable
this implies that the $$\underline\int^b_ah(x)dx = \overline\int^b_ah(x)dx$$
$$\underline\int^b_ah(x)dx = sup [{L(P,h):P ∈ P[a,b]}]>0$$ $$\underline\int^b_ah(x)dx = inf [{U(P,h):P ∈ P[a,b]}] > 0$$
since h(x) R [a,b]
$$\underline\int^b_ah(x)dx = \overline\int^b_ah(x)dx = \int^b_ah(x)dx$$
$$sup [{L(P,h):P ∈ P[a,b]}]>0 = inf [{U(P,h):P ∈ P[a,b]}] > 0$$
this means the point at which the function converges which is at the integral
$$\int^b_ah(x)dx > 0$$
I dont know based on what i did if this truly proves the question i am hoping someone can correct me i
There exists $m>0$ with $h(x)\ge m$ for all $x\in[a,b]$. This means that for any partition $P$ of $[a,b]$, the infimum of $h$ over any interval in $P$ is at least $m$. Thus $L(h,P)$, the lower integral with repsect to $P$, is at least $m(b-a)$. Since $m>0$ and $b-a>0$, this product is positive; since $h$ is integrable, the integral is positive.