# Thread: Fundamental Theorom of Calculus

1. ## Fundamental Theorom of Calculus

Hi, so I got this problem:
f(x) = x - 2√x, intervals are: [0,2]

This is what I did:
1. x - 2x1/2

Then I integrated it
2. 02(x - 2x1/2)dx

Then I used the antiderivative
3. (x2/2) - (4/3)x3/2
Just letting you know in case if you were wondering about the (4/3), since I know that it would be multiplied by (2/3) since it's the inverse of the exponent, then it was multiplied by the 2 on the outside.

Then I did f(b) - f(a), which means plugging in 2, then plugging in 0, and subtracting it from each other, though I left out the 0 part because it makes everything 0, so I only plugged in the 2, therefore I did not need to subtract
4. This is what I got:
(2)2/2 -(4/3)(2)3/2
(4/2) - (4/3)√23
2 - (4/3)√(4*2)
2 - (4/3) * (2√2)
2 - (8√2)/3

I don't really know where to go on from there.

Someone showed me this (and by the way, it's the correct answer in the back of the book as well)

But I am really confused on how they did those next steps!
I see the left side, which is f(c)(2-0), and I'm confused what that means, and where the 6 came from on the right side.
And so on.

Can someone explain to me or help me find a simplier way for me to understand?

Thanks!

2. ## Re: Fundamental Theorom of Calculus

i suspect you are not telling us the whole problem, just a part of it. your answer for the definite integral is correct, and matches what you were given:

2 - 8√2/3 = 6/3 - 8√2/3 = (6 - 8√2)/3

it looks if you're using an integral version of the mean-value theorem, that is, c is a value in (0,2) where the definite integral of f from 0 to 2 matches a rectangle of height f(c) and width 2 (the width of the interval [0,2]).

if we define $\displaystyle F(x) = \int_a^x f(t)\ dt$, for x in [a,b] then by the mean-value theorem, there is a c in (a,b) with:

$\displaystyle f(c) = F'(c) = \frac{F(b) - F(a)}{b - a}$.

that is:

$\displaystyle f(c)(b - a) = \int_a^b f(t)\ dt$

3. ## Re: Fundamental Theorom of Calculus

Originally Posted by Deveno
i suspect you are not telling us the whole problem, just a part of it. your answer for the definite integral is correct, and matches what you were given:

2 - 8√2/3 = 6/3 - 8√2/3 = (6 - 8√2)/3

it looks if you're using an integral version of the mean-value theorem, that is, c is a value in (0,2) where the definite integral of f from 0 to 2 matches a rectangle of height f(c) and width 2 (the width of the interval [0,2]).

if we define $\displaystyle F(x) = \int_a^x f(t)\ dt$, for x in [a,b] then by the mean-value theorem, there is a c in (a,b) with:

$\displaystyle f(c) = F'(c) = \frac{F(b) - F(a)}{b - a}$.

that is:

$\displaystyle f(c)(b - a) = \int_a^b f(t)\ dt$
Oh sorry!
I forgot to put in the specifics!
My bad, this is what we're trying to find:

Find the value of c guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval.
So I can see within that Mean Value Theorom, they multiplied the denominator into F'(C), right?

And I see now, so that 6 came from the 2, which made it into 6/3 so they're all on the same fraction!

Thanks

Then after that, they just divide 2 from both side since the left side had the (2-0) thing right?