Help w/First Fundamental Theorem of Calculus

[Hurricane3]

For the first fundamental theorem of calculus, I don't quite understand it...

I think I got the integral part, but not the interval [a,x]...

can you guys help me? thx

 

[SiddharthM]

Well your question is very vague but I will try and address any misunderstanding you may have.

The first fundamental theorem of calculus says that given a real valued function defined on [a,b] and if we define a new function F (note that little f and big F are two different functions here) by

F(x)=integral (lower limit is a, upper limit is x) of f(x) dx

then F(x) is continuous on all x in the closed interval [a,b]. Furthermore, if the function f is continuous at a point u that is inside the closed interval [a,b] then F is differentiable at u and it's derivative is f(u).

the first part of the theorem says that if you change the upper limit of integration a LITTLE bit then the change in integrals is also LITTLE. The second part is just saying that if f is continuous at a point then F is SMOOTH (ie differentiable) at that point and that the derivative operation and integral undo one another.

[a,x] is a subset of [a,b]. It is a fact that if f is integrable on [a,b] and x is a point in the interval [a,b] then f is integrable on [a,x].

hope this helps.

 

[Hurricane3]

ah i still dun understand...

can we work on an example from the textbook?
The example is:
Find the derivative of
\intcost dt from 0 to \sqrt{x}
by

a. evaluating the integral and differentiating the result
b. by differentiating the integral directly

for part a.
The integral of cos(t) is sin(t). Thus the integral from 0 to \sqrt{x} is
sin\sqrt{x} - sin(0)
= sin\sqrt{x}
Differentiating that, I get \frac{cos\sqrt{x}}{2\sqrt{x}}

I guess my problem is differentiating an integral directly...

What would happen if its from sin(x) to x? or from 5 to x?