the integral question again !?

Question

if there is an integral (integral(from 0 to 8) T'(X)dx),
does this simply mean the Y value when T(X) is 8?

it was derivative... there is " ' " between T and (X).
if it was just T(X) itself... I would do T(8)-T(0).
but since it is derivative... the integral(which means anti-derivative) and the " ' (which means derivative..)" cancel out..?
it was my question :) I think ' this mark was hard to see....

ah ! one more question ! when there is a table... with x values
and y values... I graphed on a calculator.. and found the
equation for the funtion too... but the question asks if
second derivative of the function of the table is always above zero... which means always concave up...
well... since I found the function of QuadReg(texas 83 says...)
and connected the "dots" of each value... I know that
it is concave up by simply looking at it.. But I think I need more
info to prove that it is concave up... ! (I guess?)
how can I prove that function is concave up...?

Answer 1

Gotta keep the faith on this one.
Since all you start with is selected x and y values, you have to accept that what you see is what you get. Make sure you understand what you mean, because I think you are getting ahead of yourself.
If you start with T(x) being linear, T'(x) is a constant and T''(x) is zero. If T"(x) is always greater than zero, T'(x) has to be always increasing, but not necessarily concave upwards.
T(x) has to be concave upwards.

Answer 2

I stick with my original answer. I did see the " ' " (prime). You're taking the integral of the function T'(x), which is the derivative of the function T(x).

∫[0→8] T'(x) = T(8) - T(0)


For the second part of the question, this is how it's done:

First, find all the critical points of the function. Use them to make a bunch of number intervals. Find the first and second derivative at any number in each of the intervals. If the sign of the first derivative is positive, the function is increasing, and if it's negative, it's decreasing. If the sign of the second derivative is positive, it's concave up. If the sign of the second derivative is negative, it's concave down.

Your question is asking if the function is always concave up. You need to find the second derivative at several different intervals to see if the sign of the 2nd derivative is positive or negative. If positive, concave up. If negative, concave down.

Answer 3

My original answer still stands, I saw the T'(x). If you take the integral of a derivative T'(x), you get back to T(x). Then to evaluate the definite integrate, you plug the limits into the function (upper limit minus lower limit). So it is T(8) - T(0).

If f"(x) > 0, then the function is concave up, so the given information proves it regardless of you knowing what the equation was anyway.

Answer 4

It is not T(8) alone.
It is T(8) - T(0).
If T(0) = 0, then it does not matter.

The definition goes as follows:
f'(x) = [f(x+dx) - f(x) ] / dx
So f(x+dx) - f(x) = f'(x) dx
So applying this over limits a and b,
int (from a to b) f'(x) dx = f(b) - f(a)

Answer 5

? dx /x *?ln x from [ a million to e^8 ] permit ln x = u: while x = a million u = 0 and while x = e^8, u = 8 dx /x = du now the imperative turns into ? du /?u from [ 0 to eight ] = ? (u)^(-a million/2) du from [ 0 to eight ] = 2?u from [ 0 to eight] = 2 [ ?8 ] = 4?2