Question

Will mark brainliest

Answer 1

$x^3$ is strictly increasing on [0, 5], so

$\max\{x^3 \mid 0\le x\le5\} = 5^3 = 125$

and

$\min\{x^3 \mid 0 \le x\le5\} = 0^3 = 0$

so the integral is bounded between

$\displaystyle \boxed{0} \le \int_0^5x^3\,dx \le \boxed{125}$