# Thread: Mean value theorem for integrals?

1. ## Mean value theorem for integrals?

Find the value of c guaranteed by the mean value theorem for integrals for the function over the indicated interval

f(x)= x-2 square root(x) [0,2]

I am not really sure how to do this as my textbook does not really provide an example. Can anyone show me how to solve it?

2. ## Re: Mean value theorem for integrals?

Are you familiar with this proposition:
If a function $f$ is continuous in $[a,b]$ and differentiable in $]a,b[$ then their exists an element $c \in ]a,b[$ wherefore:
$\int_{a}^{b} f(x)dx = (b-a)\cdot f(c)$

You've given a,b so you can integrate the function and calculate c.

3. ## Re: Mean value theorem for integrals?

I have integrated the function to get

(1/2)x^2-(4/3)x^(3/2)+c

4. ## Re: Mean value theorem for integrals?

That's indeed the primitive function. But now we have a definite integral so:
$\int_{1}^{2} (x-2\sqrt{x})dx =\int_{1}^{2}xdx - 2\int_{1}^{2}x^{\frac{1}{2}}dx=\left[\frac{x^2}{2}-\frac{4x^{\frac{3}{2}}}{3}\right]_{1}^{2}=...?$

If you've calculated this definite integral then you've:
$\int_{1}^{2} (x-2\sqrt{x})dx=(2-1)\cdot f(c)$
If $f(x)=x-2\sqrt{x}$ then $f(c)=c-2\sqrt{c}$

That means:
$\int_{1}^{2} (x-2\sqrt{x})dx=c-2\sqrt{c}$

Solve this equation for $c$

5. ## Re: Mean value theorem for integrals?

I got -1.771 when plugging in [0,2] but I am still confused on finding c. I made I typo also I mean between [0,2].

6. ## Re: Mean value theorem for integrals?

If the interval is [0,2] then:
$\int_{0}^{2} (x-2\sqrt{x})dx=2-\frac{8\sqrt{2}}{3}$ which is approximately $-1,771..$.
So:
$-1,77124=2\cdot(c-2\sqrt{c})$

Solve this equation for c.

7. ## Re: Mean value theorem for integrals?

So I have

-1.771=2(c-2 square root(c))

-1.771=2c-4c

-1.771=-2c
c=.8855 ?

Yet the back of my book says .4380 as the answer.

8. ## Re: Mean value theorem for integrals?

Originally Posted by homeylova223
So I have

-1.771=2(c-2 square root(c))

-1.771=2c-4c

-1.771=-2c
c=.8855 ?

Yet the back of my book says .4380 as the answer.
...

9. ## Re: Mean value theorem for integrals?

Hmm what would 2 times 2 square root(c) be?

10. ## Re: Mean value theorem for integrals?

Originally Posted by homeylova223
Hmm what would 2 times 2 square root(c) be?
4 square root(c)

11. ## Re: Mean value theorem for integrals?

Alright I have

2c-4 square root(c)=-1.771 I am actually having an problem solving this do I have to square the square square root c and -1.771

How would I go about solving this?

12. ## Re: Mean value theorem for integrals?

Originally Posted by homeylova223
Alright I have

2c-4 square root(c)=-1.771 I am actually having an problem solving this do I have to square the square square root c and -1.771

How would I go about solving this?

$2-\frac{8}{3}\sqrt{2}=2(c-2\sqrt{c})$

$1-\frac{4}{3}\sqrt{2}=c-2\sqrt{c}$

$1-\frac{4}{3}\sqrt{2}-c=-2\sqrt{c}$

$(1-\frac{4}{3}\sqrt{2}-c)^2=(-2\sqrt{c})^2$

$1+\frac{32}{9}+c^2-\frac{8}{3}\sqrt{2}-2c+c \frac{8}{3}\sqrt{2}=4c$

$\frac{41}{9}+c^2-\frac{8}{3}\sqrt{2}-2c+c \frac{8}{3}\sqrt{2}=4c$

$c^2+c(\frac{8}{3}\sqrt{2}-4-2)+\frac{41}{9}-\frac{8}{3}\sqrt{2}=0$

$c^2+c(\frac{8}{3}\sqrt{2}-6)+\frac{41}{9}-\frac{8}{3}\sqrt{2}=0$

Solve this for $c$.