How do you calculate the derivative of #intsin^3(2t-1) dt # from #[x, x-1]#?
1 Answer
Use the Fundamental Theorem of Calculus Part 1 and some rewriting and the chain rule. The derivative is
Explanation:
The Fundamental Theorem of Calculus, Part 1 (FTC 1) tells us that for
To find the derivative of
we note that the function is continuous on
So we will rewrite.
See properties of the definite integral for
#int_a^b f(t) dt = int_a^c f(t) dt + int_c^b f(t) dt# (Pick any constant you like where I have used
#0# .)
See properties of the definite integral for
#int_a^b f(t) dt = -int_b^a f(t) dt#
We are almost there, but the second integral doesn't just have an
For
#g(x) = int_a^u f(t) dt# , we get#g'(x) = f(u) (du)/dx#
So for this problem we arrive at:
So the derivative (with respect to
The answer can be rewritten as:
Final Note
There are people around who use the FTC 1 in the form:
The derivative of
Memorized this way, we can skip all the set-up work I did.
I've never taught from a textbook in the US that does it this way.