d/dx of the integral from 1 to 3 of cos(e^y) dy
What do the 1 and 3 mean in this case?
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.
This is a problem that requires using the second fundamental theorem of calculus, namely $\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \int_a^x{f(y)\,dy} = f(x) \end{align*}$. First note that $\displaystyle \displaystyle \begin{align*} \int_1^3{\cos{\left( e^y \right) } \,dy} + \int_3^x{\cos{\left( e^y \right) } \,dy} = \int_1^x{\cos{\left( e^y \right) } \,dy} \end{align*}$. From this we can determine
$\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \int_1^3{\cos{\left( e^y \right) } \, dy} &= \frac{d}{dx} \int_1^x{ \cos{ \left( e^y \right) }\,dy } - \frac{d}{dx} \int_3^x { \cos{ \left( e^y \right) } \,dy } \\ &= \cos{ \left( e^x \right) } - \cos{ \left( e^x \right) } \\ &= 0 \end{align*}$
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.
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)