# Math Help - derivative of an integral

1. ## derivative 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?

2. ## Re: derivative of an integral

The 1 and 3 mean the boundaries for y that you are integrating over...

3. ## Re: 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.

4. ## Re: 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.

5. ## Re: 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.

ABf(x)dx = F(B) –F(A)

d/dx∫ABf(x)dx = d/dx∫ABf(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
ABf(x,t)dt = F(x,B) – F(x,A)
d/dx∫ABf(x,t)dt = d/dxF(x,B) – d/dxF(x,A) where A and B are constants or functions of x.

d/dy∫ABf(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:
ABd/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.

6. ## Re: derivative of an integral

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

What do the 1 and 3 mean in this case?
This is a problem that requires using the second fundamental theorem of calculus, namely \displaystyle \begin{align*} \frac{d}{dx} \int_a^x{f(y)\,dy} = f(x) \end{align*}. First note that \displaystyle \begin{align*} \int_1^3{\cos{\left( e^y \right) } \,dy} + \int_3^x{\cos{\left( e^y \right) } \,dy} = \int_1^x{\cos{\left( e^y \right) } \,dy} \end{align*}. From this we can determine

\displaystyle \begin{align*} \frac{d}{dx} \int_1^3{\cos{\left( e^y \right) } \, dy} &= \frac{d}{dx} \int_1^x{ \cos{ \left( e^y \right) }\,dy } - \frac{d}{dx} \int_3^x { \cos{ \left( e^y \right) } \,dy } \\ &= \cos{ \left( e^x \right) } - \cos{ \left( e^x \right) } \\ &= 0 \end{align*}

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.

thanks!

8. ## Re: derivative of an integral

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

What do the 1 and 3 mean in this case?
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 Δxk and let the points in the interval be xk.

Then, by definition,

ABf(x)dx = lim n→∞ k=0Σk=nf(xk) Δxk

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)

ABf(x)dx = F(B) –F(A) where F’(x) = f(x)