What are the guaranteed values of c for the Mean Value Theorem for Integrals?

In summary, the Mean Value Theorem for Integrals states that the average area under a graph is equal to 1/b-a times the function's value at the given point.
  • #1
kings13
27
0

Homework Statement



Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.

Homework Equations



f(x)=3cos(x), [-pi/4, pi/4]


The Attempt at a Solution



well i do f(b)-f(a)/b-a and get zero but that's not the answer whenever i try it..
 
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  • #2
What does the Mean Value Theorem for Integrals say? (It's more specific than just the Mean Value Theorem).
 
  • #3
Strants said:
What does the Mean Value Theorem for Integrals say? (It's more specific than just the Mean Value Theorem).

the average area under the graph?

1/b-a times f(c) right? but how to put numbers into that
 
  • #4
What are a and b? What is f in this problem? What is the value of the integral, namely this one?
[tex]\int_{-\pi/4}^{\pi/4} 3 cos(x)dx[/tex]
 
  • #5
Mark44 said:
What are a and b? What is f in this problem? What is the value of the integral, namely this one?
[tex]\int_{-\pi/4}^{\pi/4} 3 cos(x)dx[/tex]

a is -pi/4

b is pi/4

f is 3cosx

integral is 0. cause odd function right?
 
  • #6
Yes, except that the function snt odd and the integral doesn't equal 0.
 
  • #7
micromass said:
Yes, except that the function snt odd and the integral doesn't equal 0.

3cos(pi/4)-3cos(-pi/4)=2.12-2.12. or is it plus? if it is plus then what next? multiply by 1/b-a?
 
  • #8
Since when does [tex]\int{3\cos(x)dx}=3cos(x)[/tex]?
What is the integral of a cosine?
 
  • #9
micromass said:
Since when does [tex]\int{3\cos(x)dx}=3cos(x)[/tex]?
What is the integral of a cosine?

not thinking straight. sorry I am doing an essay at the same time but that's really no excuse

3sin(pi/4)-3sin(pi/4)=

2.12132+2.12132


now what?
 
  • #10
Now you need to calculate

[tex]\frac{1}{b-a}\int_a^b{f(x)dx}[/tex]

thus plug in everything you know, and you will get a certain number. Now you only need to find a c such that f(x)=3cos(c) equals that very number.
 
  • #11
micromass said:
Now you need to calculate

[tex]\frac{1}{b-a}\int_a^b{f(x)dx}[/tex]

thus plug in everything you know, and you will get a certain number. Now you only need to find a c such that f(x)=3cos(c) equals that very number.

well the integral is 4.24. the 1/b-a is .63662

4.24x.63662=2.7

2.7=3cos(x)? what? how in the world do you solve that
 
  • #12
The integral is actually 3√2 and the length of the interval is actually ∏/2. You'll get better results if you leave your intermediate results in their exact forms rather than rounding to a random number of decimal places.

If you have a = b cos(x), there's an inverse trig function that will be helpful.
 
  • #13
inverse cosine i got it right

thanks for the help
 
  • #14
Mark44 said:
The integral is actually 3√2 and the length of the interval is actually ∏/2. You'll get better results if you leave your intermediate results in their exact forms rather than rounding to a random number of decimal places.

If you have a = b cos(x), there's an inverse trig function that will be helpful.

the online thing wanted decimals but yea i shud leave it as that for the test.

before someone calls me dumb for not using inverse trig i solved it before reading the last help
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a mathematical concept that states that if a function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one point c in (a,b) where the instantaneous rate of change (derivative) of the function at c is equal to the average rate of change of the function over the interval [a,b].

2. What is the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals is a special case of the Mean Value Theorem that applies to integrals. It states that for a continuous function f(x) on the closed interval [a,b], there exists at least one point c in (a,b) where the value of the integral from a to b is equal to the function evaluated at c multiplied by the length of the interval (b-a).

3. How do you use the Mean Value Theorem to solve integrals?

To use the Mean Value Theorem to solve integrals, you first need to identify the function f(x) on the given interval [a,b]. Then, you can use the Mean Value Theorem for Integrals to set up an equation where the integral of f(x) from a to b is equal to f(c) multiplied by (b-a), where c is the unknown point on the interval. You can then solve for c to find the specific value of f(c) that satisfies the equation.

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

The Mean Value Theorem for Integrals allows us to find the exact value of a definite integral without having to evaluate it using traditional methods such as the Fundamental Theorem of Calculus or Riemann sums. It also has many real-world applications, such as in physics and economics, where finding the average value of a function over a certain interval is important.

5. Are there any limitations to the Mean Value Theorem for Integrals?

Yes, there are a few limitations to the Mean Value Theorem for Integrals. First, the theorem only applies to continuous functions, so it cannot be used for functions with discontinuities or sharp turns. Additionally, the theorem only guarantees the existence of at least one point c, but it does not provide a method for finding the exact value of c. Finally, the theorem does not apply to improper integrals, as they do not have a finite interval over which the function is defined.

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