$f(x) = \frac{d}{dx} \int_{x^3}^{x^2} (t^2 +1) dt$, Find the explicit expression of $f(x)$
$\int_{x^3}^{x^2} (t^2 +1) dt = \int_{x^3}^0 (t^2 +1) dt + \int_{0}^{x^2} (t^2 +1) dt = \int_0^{x^2} (t^2+1) dt -\int_0^{x^3} (t^2+1) dt$
However, if I solve the integral directly: $\int_{x^3}^{x^2} (t^2 +1) dt = \frac{1}{3}x^6 + x^2 - \frac{1}{3}x^9 - x^3$
However, if I solve $\int_0^{x^2} (t^2+1) dt -\int_0^{x^3} (t^2+1) dt = \frac{1}{3} x^9 +x^2- \frac{1}{3}x^6 - x^3$
Both answers is different, where have I gone wrong in my first working step?
Disclaimer: I know how to solve this by applying the first fundamental theorem of calculus, and now for learning purposes, when I am try to integrate the function and then differentiate it, I made an error. Where have I gone wrong in my definite integral manipulation of $\int_{x^3}^{x^2} (t^2 +1) dt = \int_{x^3}^0 (t^2 +1) dt + \int_{0}^{x^2} (t^2 +1) dt = \int_0^{x^2} (t^2+1) dt -\int_0^{x^3} (t^2+1) dt$
