Fundamental Concept of the Indefinite VS Definite Integral

unders

New member
Joined
Jan 29, 2012
Messages
2
I've learned AP calculus BC (got score 5) and likely can do most non-differenctial equation calculus questions. Yet, I woke up one day surprised to find out that I have no idea what the right answer is to the very fundamental concept below. If you know the right answer, or can cite a calculus theorem or definition to address the follow, please do.

We know:

calculus 3.jpg


Thank you.
 
Last edited:
I've learned AP calculus BC (got score 5) and likely can do most non-differenctial equation calculus questions. Yet, I woke up one day surprised to find out that I have no idea what the right answer is to the very fundamental concept below. If you know the right answer, or can cite a calculus theorem or definition to address the follow, please do.

We know:


View attachment 1652
I honestly never understood this. Someone please give me a correct answer that has a Yes or No in it.

Thank you.

Yes

\(\displaystyle \int_0^kf(x) dx \ = \ \int f(x)dx |_0^k \ = \ F(x) |_0^k\)
 
I want to double check:

If i plug K into the antiderivative (evaluate the antiderivative at k), is it ALWAYS EQUAL to the area from underneath the original derivative from 0 to k???? What about the constant + C?
 
Last edited:
I want to double check:

If i plug K into the antiderivative (evaluate the antiderivative at k), is it ALWAYS EQUAL to the area from underneath the original derivative from 0 to k???? What about the constant + C?

No... Even without considering constants.

\(\displaystyle \displaystyle \int_0^{\pi/2} \sin(x) dx = -\cos(x)\bigg\vert_{0} ^{\pi/2} = 1 \neq 0 = -\cos(\pi/2)\).

In addition, the function should be non-negative for the whole "area" thing to be valid.
 
Top