# help on prove a integral is bigger than 0

• Nov 20th 2009, 02:02 PM
yht0251
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.
• Nov 20th 2009, 02:09 PM
Drexel28
Quote:

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?
• Nov 20th 2009, 02:16 PM
yht0251
Quote:

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?
• Nov 20th 2009, 02:18 PM
Drexel28
Quote:

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.
• Nov 20th 2009, 02:21 PM
yht0251
Quote:

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?
• Nov 20th 2009, 02:22 PM
Scott H
Is $\displaystyle p$ continuous?

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

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

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

What we need to do is find a way to show that there is some positive area under the curve $\displaystyle y=p(x)$. What happens when we let $\displaystyle \epsilon=\frac{L}{2}$?
• Nov 20th 2009, 02:24 PM
Drexel28
Quote:

Originally Posted by Scott H
Is $\displaystyle p$ continuous?

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

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

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

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

what's your point? Can't the function theoretically be integrable and not continuous?
• Nov 20th 2009, 02:25 PM
Drexel28
Quote:

Originally Posted by Scott H
Is $\displaystyle p$ continuous?

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

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

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

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

what's your point? Can't the function theoretically be integrable and not continuous?

@OP-what class are you in?
• Nov 20th 2009, 02:34 PM
yht0251
Quote:

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.