Mean value theorem for integrals

In summary: Theorem:If $f$ is continuous on $[a,b]$, then there exists $c\in [a,b]$ such that$$f(c)= \frac{1}{b-a} \int_{a}^{b}f(x) \, dx.$$
  • #1
Yankel
395
0
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!
 
Physics news on Phys.org
  • #2
Yankel said:
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

$\chi$ $\sigma$
 
  • #3
View attachment 1160

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]?
 

Attachments

  • yankel.png
    yankel.png
    2 KB · Views: 81
  • #4
chisigma said:
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

$\chi$ $\sigma$
 
  • #5
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?
 

FAQ: Mean value theorem for integrals

What is the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals is a fundamental theorem in calculus that states that if a function is continuous on a closed interval, then there exists at least one point within that interval where the average value of the function is equal to the value of the function at that point.

What is the significance of the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals is significant because it allows us to find the average value of a function over a given interval, which can be useful in various applications, such as determining the average velocity of an object or the average rate of change of a quantity over time.

How is the Mean Value Theorem for Integrals related to the Mean Value Theorem for Derivatives?

The Mean Value Theorem for Integrals and the Mean Value Theorem for Derivatives are both based on the same concept, but they apply to different types of functions. The Mean Value Theorem for Derivatives applies to differentiable functions, while the Mean Value Theorem for Integrals applies to continuous functions. However, both theorems state that there exists at least one point where the average rate of change or average value is equal to the instantaneous rate of change or value at that point.

How do you prove the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals can be proven using the Intermediate Value Theorem and the Fundamental Theorem of Calculus. The proof involves showing that the average value of the function over the interval is equal to the value of the function at a specific point within the interval, which is then shown to be true using the Intermediate Value Theorem.

What are the applications of the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals has various applications in physics, economics, and other fields of science. It can be used to find the average velocity of an object, the average rate of change in a quantity over time, and the average value of a function over a given interval. It can also be used in optimization problems to find the maximum or minimum average value of a function.

Similar threads

Back
Top