Welcome to our community

Be a part of something great, join today!

Mean value theorem for integrals

Yankel

Active member
Jan 27, 2012
398
Hello all,

I have a couple of questions.

First, about the mean value theorem for integrals. I don't get it. The theorem say that if f(x) is continuous in [a,b] then there exist a point c in [a,b] such that

\[\int_{a}^{b}f(x)dx=f(c)\cdot (b-a)\]

Now, I understand what it means (I think), but don't get the intuition (unlike the mean value theorem which is very intuitive). The integral is the area under f(x) between a and b. So how come it is equal to the height of f(x) at the point c, multiplied by the width ? How come there is a point c that represent the "average height" ?

The second question, I need to approximately evaluate

\[\int_{0}^{2}e^{x^{2}}dx\]

the answer is [2,2e^4], don't know why...

Thanks!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello all,

I have a couple of questions.

The second question, I need to approximately evaluate

\[\int_{0}^{2}e^{x^{2}}dx\]

the answer is [2,2e^4], don't know why...
The function $\displaystyle y=e^{x^{2}}$ doesn't have an elementary antiderivative, but a way to solve the integral is to use the Taylor expansion...


$\displaystyle e^{x^{2}} = \sum_{n=0}^{\infty} \frac{x^{2 n}}{n!}\ (1)$

... and from (1) You obtain...


$\displaystyle \int_{0}^{\xi} e^{x^{2}}\ dx = \sum_{n=0}^{\infty} \frac{\xi^{2 n + 1}}{(2n + 1)\ n!}\ (2)$


... and a 'good' value of the integral can be obtained summin a 'large enough' number of terms of (2)...

Kind regards

$\chi$ $\sigma$
 

M R

Active member
Jun 22, 2013
51
yankel.png

My diagram is intended to show that [tex]\displaystyle f(0)\times (2-0)< \int_0^2 e^{x^2}dx < f(2) \times (2-0)[/tex]

Can you see that [tex]\displaystyle \int_0^2 e^{x^2}dx = f(c) \times (2-0)[/tex] for some c in [0,2]?
 

chisigma

Well-known member
Feb 13, 2012
1,704
The function $\displaystyle y=e^{x^{2}}$ doesn't have an elementary antiderivative, but a way to solve the integral is to use the Taylor expansion...


$\displaystyle e^{x^{2}} = \sum_{n=0}^{\infty} \frac{x^{2 n}}{n!}\ (1)$

... and from (1) You obtain...


$\displaystyle \int_{0}^{\xi} e^{x^{2}}\ dx = \sum_{n=0}^{\infty} \frac{\xi^{2 n + 1}}{(2n + 1)\ n!}\ (2)$


... and a 'good' value of the integral can be obtained summin a 'large enough' number of terms of (2)...
Using 'Monster Wolfram' You obtain...

$\displaystyle \int_{0}^{2} e^{x^{2}}\ dx \sim 16.4396\ \text{with 10 terms}$

$\displaystyle \int_{0}^{2} e^{x^{2}}\ dx \sim 16.4515\ \text{with 12 terms}$

$\displaystyle \int_{0}^{2} e^{x^{2}}\ dx \sim 16.4526\ \text{with infinite terms}$

Kind regards

$\chi$ $\sigma$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
If you rearrange the MVT for integrals, you get that if $f$ is continuous on $[a,b]$, then there exists $c\in [a,b]$ such that
$$f(c)= \frac{1}{b-a} \int_{a}^{b}f(x) \, dx.$$
In words, $f$ must achieve (LHS) its average value on $[a,b]$ (RHS). Because $f$ is continuous, it's not allowed to skip its average value on the interval. Does that help with the intuition?