Question

using the Mean Value Theorem, find the integral of square root of x with the values of [4,9], i have the formula but, i am gettin lost with the formula

Answer 1

\[F \prime = \frac{ f(b)-Fa) }{ b-a }\]
just substitute
and their conditions is met with this interval

Answer 2

b = 9 a = 4
but are you sure the question asked you about the integration??

Answer 3

sorry, umm the formula is \[\int\limits_{a}^{b}f(x)dx=f(c)[b-a]\]

Answer 4

\[5\sqrt{x}\]

Answer 5

To clarify... you want to evaluate the integral
\[\int_4^9\sqrt x~dx\]
using the MVT, which says if a function is continuous in the closed interval [a,b] and differentiable in the open interval (a,b), then there is some c between a and b such that
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]
In this case (for the integral MVT),
\[f(b)-f(a)=\int_a^b f'(x)~dx~~\iff~~f(9)-f(4)=\int_4^9\sqrt x~dx\]

Answer 6

but that isn't the MVT at all dx doesn't equal 5 !!!
MVT is the same as linear approximation except it gives an exact value .

Answer 7

The MVT you're familiar with (involving derivatives) is the same for integrals with a slight variation. I'm not sure what you're trying to say @Catch.me

Answer 8

yeah he is right

Answer 9

what i meant that\[dx \neq b-a =\]
look at the last formula in that link
http://www.cut-the-knot.org/Curriculum/Calculus/MVT.shtml

Answer 10

No one's claiming that \(dx=b-a\).

The question itself is kinda weird to begin with, @nutterbutter. The MVT says
\[F'(c)=\frac{F(b)-F(a)}{b-a}~~\Rightarrow~~f(c)=\frac{1}{b-a}\int_a^b f(t)~dt\]
In order to find the integral, you need the average value, but to find that you need to know the integral... See the problem here?

Answer 11

OK
and what he asked is to find the integration by MVT,but you wrote \[\int\limits_{a}^{b}\sqrt{x} dx =F(9) - F(4)\]
which isn't due to MVT
but what related is that
\[\int\limits_{a}^{b}\sqrt{x} dx = f(c)(b-a).= 5 \sqrt{x}.\]

Answer 12

or I got you wrong I am sorry.
I think the answer is 5 sqrt x

Answer 13

i meant what sithsandgiggles is saying

Answer 14

The answer must be a number. A definite integral is always a number.

Answer 15

OK then you will integrate \[\sqrt{X}\] to get F(b) - F(a) right??

Answer 16

yes

Answer 17

Thus you didn't use MVT can you understand me ??
MVT didn't proved F(b)-F(a) = integral f(x).
,so your methods won't use MVT and that is my point MVT said
integral/b-a = f(c) -----> integral ={b-a}f(c).

Answer 18

.........

Answer 19

you didn't get me yet ??

Answer 20

This is what I mean MVT says about the integration that:
\[\int\limits_{a}^{b} f(x)dx = \left[ b-a \right] f(c).\]