PLEASE HELP! MEDAL! @hero @mathmale @jim_thompson5910 @SolomonZelman
@iGreen @satellite73 @SolomonZelman @xapproachesinfinity
\(\color{#000000}{\displaystyle\int\limits_{4}^{5} (-35)~dx}\) (The indefinite integral of -35 is just -35x) \(\color{#000000}{\displaystyle\left. -35x {\color{white}{\Large |}}\right|_{4}^{5} =(-35\times 5)-(-35\times 4)=-175+140=-35}\)
This is an example, and I did it doing calculus, although you can use geometry too for constant functions
so, then it'll be -20, right?
No....
Why did you think that?
-20x?
yes, that would be the indefinite integral.
And in a case of your definite integral, your evaluating -20x from x=3 to x=7
Oh, okay. I don't completely understand how you get that though
-20x I mean
We can prove why derivative of ax (with respect to x) is just a, and then you know that integration and differentiation are operations that are reverses of each other.
would you like to prove this, or ... ?
So we're just finding the antiderivative?
Yes, the integral is antiderivative. And in case when you have the limits of integration, then you are finding the derivative and then evaluating that at x=7 and x=3, and subtract.
Oh, ok. Integral notation always confuses me
The theory is: So suppose that \(\color{#000000}{\displaystyle\int\limits_{~}^{~} f(x)~dx=g(x)\color{grey}{+C}}\) then, \(\color{#000000}{\displaystyle\int\limits_{a}^{b} f(x)~dx=g(b)-g(a)}\)
is this making sense?
(as short and as precise as I can write it)
yes
Thank you. :)
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