Question

Integral with variable bounds?

Answer 1

I'm a little rusty on this, but let me see if I can help:

For an integral with an upper limit of integration as a variable, we take the variable limit and put that into our function f(t). This is because the upper limit of integration is a function itself. When you have an integral like that, what you're really looking at is a function within a function or h(g(x))

let g(x) = 2x

and h(x) = integral f(2x) dx


so your integral is just h(g(x)) which will look like this:

\[h(g(x)) = \int\limits_{3}^{g(x)} f(2x) dx\]

Notice that the question is asking you to find the derivative: since you're asked to find the derivative, just differentiate h(g(x)). Which means we have to chain rule:

d/dt (h(g(x)) = h'(g(x))*g'(x):

so we just have to figure out what h'(g(x)) is and what g'(x)

h'g(x) = d/dx (integral f(t) dt

By the Fundamental Theorem of Calculus, differentiation and integration are inverse processes, which means they cancel each other out. so the derivative of an integral is just the function itself:

\[d/dx \int\limits_{}^{} f(2x) dt = f(2x)\]

now g'(x) is easy. we said g(x) = 2x so g'(x) = 2. Which means our h'(g(x))*g'(x) is just:

f(2x)*(2)

Answer 2

Haha I think I missed the day I was taught this, but I tried my best to cobble together the answer. Sorry if its confusing :(

Answer 3

Don't worry, I get the concept :)