# Thread: need help integral theorems

1. ## need help integral theorems

State true and false
1>If f and g are contineous and f(x)>= g(x) for a<x<b, then integral(a to b)f(x)dx>=integral(a tob)g(x)dx
2>If f and g are contineous and f(x)>= g(x) for a<x<b, then f '(x)>=g'(x) for a<x<b

2. #1 is a theorem. You should see if you could prove it. It is actually very important.

#2: On [0,1] consider $\displaystyle f(x) = 2 - x^2 \quad \& \quad g(x) = x^2 .$

3. Hello, Gracy!

State true and false.

(1) If $\displaystyle f$ and $\displaystyle g$ are continuous and $\displaystyle f(x) \geq g(x)$ for $\displaystyle a<x<b$,
then: .$\displaystyle \int^b_a f(x)\,dx\:\geq \:\int^b_ag(x)\,dx$

I would say True . . .
Code:
        |   *
|   :*        :
|   :  *      :
|   :     *   :
|   :         * f(x)
|   :         :
|   :         o g(x)
|   :      o  :
|   :  o      :
|   o         :
--+---+---------+--
|   a         b

If $\displaystyle f(x)$ is always "above" $\displaystyle g(x)$ on $\displaystyle (a,\,b)$
. . then the area under $\displaystyle f(x)$ will be $\displaystyle \geq$ the area under $\displaystyle g(x).$

(2) If $\displaystyle f$ and $\displaystyle g$ are continuous and $\displaystyle f(x) \geq g(x)$ for $\displaystyle a<x<b$,
then: .$\displaystyle f '(x) \geq g'(x)$ for $\displaystyle a<x<b$

This is False . . .
Code:
        |   *
|   :*        :
|   :  *      :
|   :     *   :
|   :         * f(x)
|   :         :
|   :         o g(x)
|   :      o  :
|   :  o      :
|   o         :
--+---+---------+--
|   a         b

While $\displaystyle f(x)$ is above $\displaystyle g(x)$, we see that
. . $\displaystyle f'(x)$ is negative while $\displaystyle g'(x)$ is positive.