Fundamental Theorem of Calculus (university level)

Can someone please explain this one? Why does LHS = RHS?

\displaystyle \frac{\partial}{\partial v} \int^v_u K(t, w) \mathrm{d}t = K(v, w)

where u, v and w are functions of x.

I know it's by the fundamental theorem, but I just don't get how it's done.

Reply 1

TheCurlyHairedDude

You got happy because you got a response to your thread didn't you :smile:

Reply 2

Swayum

Original post by TheCurlyHairedDude

You got happy because you got a response to your thread didn't you :smile:

No, I had a feeling it'd be a troll judging by your username, the time and the date.

Reply 3

username425364

Original post by Swayum

No, I had a feeling it'd be a troll judging by your username, the time and the date.

I've seriously no idea how you guys keep this stuff in your head though!?
I mean, if that's not one of the hardest things ever, i'm not sure what is.

Reply 4

nuodai

The Fundamental Theorem of Calculus states that

\dfrac{d}{dx} \displaystyle \int^x_{x_0} f(t)\, dt = f(x)

, where

x_0

is some constant. Here, since we're doing a partial derivative, we can write

f(t) = K(t,w)

for a particular value of

w

(which we're holding constant). Then if

F(v) = \displaystyle \int_w^v f(t)\, dt

, the partial derivative for the original integral (which depends on both v and w) is reduced to a total derivative of

F

(which has w held constant), and so the derivative is

f(v)

. Since this is true for all values of w, this means the derivative is just equal to

K(v,w)

Reply 5

Swayum

Original post by super.teve

I've seriously no idea how you guys keep this stuff in your head though!?
I mean, if that's not one of the hardest things ever, i'm not sure what is.

The notation makes it look a lot harder than it really is, to be honest. Imagine a 10 year old walking into a GCSE maths class - he'd find it impossible to comprehend, but in reality every 16 year old in UK is expected to be able to do it.

Reply 6

username425364

Original post by Swayum

No seriously, this is out of my league, I can't comprehend it, I will never and I can't even begin to even look at it.
You math dudes are serious hardcore absolute geniuses.

Reply 7

Swayum

Original post by nuodai

The Fundamental Theorem of Calculus states that

\dfrac{d}{dx} \displaystyle \int^x_{x_0} f(t)\, dt = f(x)

, where

x_0

is some constant. Here, since we're doing a partial derivative, we can write

f(t) = K(t,w)

for a particular value of

w

(which we're holding constant). Then if

F(v) = \displaystyle \int_w^v f(t)\, dt

, the partial derivative for the original integral (which depends on both v and w) is reduced to a total derivative of

F

(which has w held constant), and so the derivative is

f(v)

. Since this is true for all values of w, this means the derivative is just equal to

K(v,w)

Ah ok, thanks that makes sense. Do we not need to multiply by dv/dx as well though? Like, if we had

\dfrac{d}{dx} \displaystyle \int^{g(x)}_{x_0} f(t)\, dt

, then we should get f(g(x))g'(x) right?

Reply 8

Farhan.Hanif93

Original post by Swayum

Ah ok, thanks that makes sense. Do we not need to multiply by dv/dx as well though? Like, if we had

\dfrac{d}{dx} \displaystyle \int^{g(x)}_{x_0} f(t)\, dt

, then we should get f(g(x))g'(x) right?

Why would you multiply by dv/dx when there is no mention of a variable, x, in the original integral?
Also, v isn't a function, it's a variable so there wouldn't be a situation where you'd need to multiply by any additional derivatives. That's what makes it different to the example you've just posted, where the upper limit is g(x) i.e. a function.

(edited 13 years ago)

Reply 9

Swayum

Original post by Farhan.Hanif93

As stated in the original post, u, v and w are functions of x...

I think the point is that you're multiplying by dv/dv = 1, so there's no point writing it, but I wanted confirmation.

Reply 10

SimonM

Original post by Swayum

As stated in the original post, u, v and w are functions of x...

I think the point is that you're multiplying by dv/dv = 1, so there's no point writing it, but I wanted confirmation.