Determine the intervals where f(x) is increasing

Question

So I have this function here. I am supposed to find the intervals of increase for this and I just want to check I did it correctly

$f(x)=\int_{0}^{\sinh(x)} (1-t^2)(e^{t^3}-1) dt $

I then applied the fundamental theorem of calculus to get

$f'(x)=(1-\sinh^2(x))(e^{\sinh^3(x)}-1)\cosh(x) $

If I find the critical points, I get $x=\sinh^{-1}(1),-\sinh^{-1}(1),0$

If I apply the first derivative test, I get that it is increasing from $(-\infty,-\sinh^{-1}(1))\cup(0,\sinh^{-1}(1))$ Since the derivative is positive on those intervals. Is this correct?

EDIT: My only issue is my domain of consideration. Do I have to consider points below 0? The bounds of the integral start from 0 and go to $\sinh(x)$ so I'm not sure if that's something I have to consider.

Answer 1

First of all, two of your critical points are incorrect. From the derivative you do get that one of the solutions comes from $\sinh(x)=1$, but that does NOT imply that "$x=\sinh(1)$"; instead the correct solution is $x=\sinh^{-1}(1)$ (the inverse function). Same for the one with $-1$. The critical point $x=0$ is correct.

To answer you question in the end of your post: yes, you must consider all reals as the domain for $x$. A definite integral can have its upper limit of integration to be a number less than its lower limit. It's interpreted according to the property: $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$.

Answer 2

$$\int_a^bf(t)dt=-\int_a^bf(t)dt$$ Just because the top part is less than the bottom part does not make it invalid. Unless you have some reason to not consider $(-\infty, -\sinh(1))$ as part of your domain, you should consider it a part of your domain.