F is Continuous and Nonnegative the Integral is 0 the Function is 0 Chegg
Prove that if $f$ is continuous and nonnegative in the interval $[a,b]$, then the integral is greater than or equal to $0$
Solution 1
Without loss of generality, assume that $[a,b]=[0,1]$. Because $f$ is Riemann integrable, by the definition you can easily deduce that $A=\lim_{n\rightarrow\infty}\displaystyle\sum_{k=1}^{n}f\left(\dfrac{k}{n}\right)\dfrac{1}{n}$. Now $f(k/n)\geq 0$ so $\displaystyle\sum_{k=1}^{n}f\left(\dfrac{k}{n}\right)\dfrac{1}{n}\geq 0$, then so is its limit.
Solution 2
Let $m$ be the minimum value and $M$ the maximum value of the continuous function $f$ on the interval $[a,b]$. Then $$ m(b-a)\le\int_a^b f(x)\,dx\le M(b-a) $$
More generally, for a Riemann integrable function $f$ over $[a,b]$, if $l$ is a lower bound for $f$, then $$ l(b-a)\le \int_a^b f(x)\,dx $$ Just take a Riemann sum relative to the trivial subdivision.
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Comments
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Prove that if $f$ is continuous and nonnegative in the interval $[a,b]$, then $$A=\int_a^b f(x) \, dx \ge 0$$
My attempted proof: Suppose otherwise i.e. $$\int_a^b f(x) \, dx < 0$$ Then by definition $\exists \delta$ such that $\forall \delta$-fine subdivisions of $[a,b]$ and any choice of $\xi_i[x_{i-1},x_i]$ then $$\left|\sum_{i=0}^n f(\xi_i)\Delta x_i-A\right|<0<\epsilon$$ This implies that $$\sum_{i=0}^n f(\xi_i)\Delta x_i-A$$ is negative.
I'm stuck at this point, any help? Also, is it possible to prove this directly? I think (I could be wrong) it's somewhat trivial since the function is nonnegative on $[a,b]$ i.e. $f(x)=0$ or $f(x)>0$ then then it should just follow that the integral would be just that.
UPDATE: John Don pointed out how I was defining the integral. I am not using Darboux definition (unfortunately) but the following:
Let$f(x)$ be a function on $[a,b]$. We say that $\int_a^b f(x) \, dx$ exists and equals A if $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall$ subdivisions of [a,b] which are $\delta$-fine (i.e. $\Delta x_i < \delta$, $\forall i$) and $\forall \xi_i\in [x_{i-1},x_i]$, then $$\left|\sum_{i=0}^n f(\xi_i)\Delta x_i-A\right|<\epsilon$$
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Any lower sum is non negative... proof done.
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Isn't it immediate from the fact that for any partition, each term in the sum is nonnegative?
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Is this part of what they want to prove? A=∫ b a f(x)dx Or just A≥0?
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$| \sum_{i=0}^n f(\epsilon_i)\Delta x_i -A |<0$ really has no sense since the left side is $\ge 0$ by definition. Now if you want to demonstrate that maybe you could use the fact that $f$ has a minimum in $[a,b].$
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@user78090 $A \ge 0$ because the former already follows from the assumptions
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If integration is area under a curve and the curve is never negative... If you define integration via a lim of sum of area slices each slice is f(x)dx in area and f(x) >=0 so the sum of these areas is non-negative.
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How have you defined the integral - if you have done it via the Draboux integral, then the statement is just trivial (assuming that you know that continuous functions are integrable).
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This makes sense.
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But wouldn't your last step just be sufficient?
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@ContraModernistae Yes, it would. But you don't really need Riemann sums for integrals of continuous functions and I don't know what you're precisely interested in.
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I see and to show that the integral nonnegative.
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@ContraModernistae In your setting, a lower bound for $f$ is $0$.
Recents
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Source: https://9to5science.com/prove-that-if-f-is-continuous-and-nonnegative-in-the-interval-a-b-then-the-integral-is-greater-than-or-equal-to-0
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