# Thread: help on prove a integral is bigger than 0

1. ## help on prove a integral is bigger than 0

hi, how can i prove the attached function is bigger than 0, if p(x) not equal to 0.

2. Originally Posted by yht0251
hi, how can i prove the attached function is bigger than 0, if p(x) not equal to 0.
Note that your integrand is always positive on that interval. I mean, depending on the context of the question I would think that'd be enough. If your integral diverged it would approach....what? Can you tell?

3. Originally Posted by Drexel28
Note that your integrand is always positive on that interval. I mean, depending on the context of the question I would think that'd be enough. If your integral diverged it would approach....what? Can you tell?
why positive integrand gives positive integral?

4. Originally Posted by yht0251
why positive integrand gives positive integral?
Non-rigorously because if a curve is above the x-axis then the area below it will be positive.

5. Originally Posted by Drexel28
Non-rigorously because if a curve is above the x-axis then the area below it will be positive.
can it be proved in a mathematical way?

6. Is $p$ continuous?

Since $p(x)^2$ doesn't distinguish between the sign of $p(x)$, we may assume that $p(x)$ is nonnegative everywhere. Now, let's suppose that $c$ is a value at which $p(c)=L\ne 0$. By the assumed continuity of $p$, we have, from the epsilon-delta definition of limit,

For all $\epsilon$, there exists a $\delta$ such that $0<|x-c|<\delta$ implies $|f(x)-L|<\epsilon$.

This just states that $\lim_{\small x\rightarrow c}f(x)=f(c)$.

What we need to do is find a way to show that there is some positive area under the curve $y=p(x)$. What happens when we let $\epsilon=\frac{L}{2}$?

7. Originally Posted by Scott H
Is $p$ continuous?

Since $p(x)^2$ doesn't distinguish between the sign of $p(x)$, we may assume that $p(x)$ is nonnegative everywhere. Now, let's suppose that $c$ is a value at which $p(c)=L\ne 0$. By the assumed continuity of $p$, we have, from the epsilon-delta definition of limit,

For all $\epsilon$, there exists a $\delta$ such that $0<|x-c|<\delta$ implies $|f(x)-L|<\epsilon$.

This just states that $\lim_{\small x\rightarrow c}f(x)=f(c)$.

What we need to do is find a way to show that there is some positive area under the curve $y=p(x)$. What happens when we let $\epsilon=\frac{L}{2}$?
what's your point? Can't the function theoretically be integrable and not continuous?

8. Originally Posted by Scott H
Is $p$ continuous?

Since $p(x)^2$ doesn't distinguish between the sign of $p(x)$, we may assume that $p(x)$ is nonnegative everywhere. Now, let's suppose that $c$ is a value at which $p(c)=L\ne 0$. By the assumed continuity of $p$, we have, from the epsilon-delta definition of limit,

For all $\epsilon$, there exists a $\delta$ such that $0<|x-c|<\delta$ implies $|f(x)-L|<\epsilon$.

This just states that $\lim_{\small x\rightarrow c}f(x)=f(c)$.

What we need to do is find a way to show that there is some positive area under the curve $y=p(x)$. What happens when we let $\epsilon=\frac{L}{2}$?
what's your point? Can't the function theoretically be integrable and not continuous?

@OP-what class are you in?

9. Originally Posted by Drexel28
Non-rigorously because if a curve is above the x-axis then the area below it will be positive.
thank you , i think i got the point. integral is some kind of adding , when everything added is positive, the sum is positive.