d/dx of the integral from 1 to 3 of cos(e^y) dy

What do the 1 and 3 mean in this case?

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- Apr 30th 2013, 09:01 PMkingsolomonsgravederivative of an integral
d/dx of the integral from 1 to 3 of cos(e^y) dy

What do the 1 and 3 mean in this case? - Apr 30th 2013, 10:52 PMProve ItRe: derivative of an integral
The 1 and 3 mean the boundaries for y that you are integrating over...

- Apr 30th 2013, 11:36 PMJJacquelinRe: derivative of an integral
Hi kingsolomonsgrave !

Do not try to find the integral from 1 to 3 of cos(e^y) dy : You cannot find it.

But you can answer the question without knowing the integral. - May 1st 2013, 08:22 AMkingsolomonsgraveRe: derivative of an integral
Is this correct? If the integral is a number when the limits of integration are numbers then the derivative of that integral is zero.

- May 1st 2013, 12:12 PMHartlwRe: derivative of an integral
The OP has already been answered. To satisfy my curiousity I pursued extensions from the point of view of the fundamental definition and pass it on.

Definition: ∫f(x)dx = F(x) where F’(x) = f(x)

d/dx∫f(x)dx = f(x) by definition.

∫_{AB}f(x)dx = F(B) –F(A)

d/dx∫_{AB}f(x)dx = d/dx∫_{AB}f(t)dt = d/dxF(B) – d/dxF(A) where F’(t) = f(t)

∫f(x,t)dt = F(x,t) + g(x) where d/dtF(x,t) = f(x,t) because d/dt[F(x,t) + g(x)] = f(x,t) and you get F(x,t) by integrating with x fixed. Then

∫_{AB}f(x,t)dt = F(x,B) – F(x,A)

d/dx∫_{AB}f(x,t)dt = d/dxF(x,B) – d/dxF(x,A) where A and B are constants or functions of x.

d/dy∫_{AB}f(x,t)dt = d/dyF(x,B) – d/dyF(x,A) where A and B are constants or functions of y.

d/dx∫f(x,t)dt = d/dxF(x,t) + g’(x) where d/dtF(x,t) = f(x,t)

∫d/dxf(x,t)dt = d/dxF(x,t) + g’(x) because d/dt(d/dx)F(x,t) + g’(x) = d/dx(d/dt)F(x,t) = d/dxf(x,t)

so d/dx∫f(x,t) = ∫d/dxf(x,t)dt if you can interchange order of differentiation of F(x,t).

Finally:

∫_{AB}d/dxf(x,t)dt = d/dx∫_{ABf}(x,t)dt = = d/dxF(x,B) – d/dxF(x,A) where d/dt F(x,t) = f(x,t)

Of course sub and superscripting were not carried over from Word so I contented myself with ∫_{AB}for integral from A to B.

Interesting example of how far you can go just from the fundamental definition. - May 1st 2013, 06:15 PMProve ItRe: derivative of an integral
This is a problem that requires using the second fundamental theorem of calculus, namely . First note that . From this we can determine

This makes perfect sense considering that when you integrate a function between two points, you get a numerical value, and the derivative of a number is 0. - May 2nd 2013, 05:30 AMkingsolomonsgraveRe: derivative of an integral
thanks!

- May 2nd 2013, 07:08 AMHartlwRe: derivative of an integral
As to the meaning of the limits of integration 1,3.

Given f(x).

For x between A and B divide the interval [A,B] into n subintervals Δx_{k}and let the points in the interval be x_{k}.

Then, by definition,

∫_{A}^{B}f(x)dx = lim n→∞_{k=0}Σ_{k=n}f(x_{k) }Δx_{k}

This is usually illustrated as the area under the curve f(x) between A and B.

It then turns out that (see a calculus text)

∫_{A}^{B}f(x)dx = F(B) –F(A) where F’(x) = f(x)