Question

Find F'(x) if:

Answer 1

\[F(x) = \int\limits_{1}^{x^3}\arcsin(t).dt\]

Answer 2

If it were simply x as the upper bound it would be arcsin(x)
Would the cube change this part of the FTC?

Answer 3

Looking at The Second Fundamental Theorem of Calculus; can anyone check if this is correct? http://ltcconline.net/greenl/courses/105/antiderivatives/secfund.htm

let \[u=x^3\]
\[y=\int\limits_{1}^{x^3}\arcsin(t).dt\]


\[\frac{ dy }{ dx } =\frac{ dy }{ du }\frac{ du }{ dx } = (\arcsin(u))(3x^2)= 3x^2\arcsin(x^3)\]

Answer 4

I would do the same thing, but I have not seen this problem before.

Answer 5

Checking the notes on your link, this is correct.