\[ I_y = \int_0^a \int_0{kx2} x^2 \, dy \, dx \] Calculando: \[ I_y = \int_0^a x^2 \left[ y \right]_0{kx2} dx \] \[ I_y = \int_0^a x^2 (kx^2) \, dx \] \[ I_y = k \int_0^a x^4 \, dx \] \[ I_y = k \left[ \frac{x^5}{5} \right]_0^a \] \[ I_y = \frac{k a^5}{5} \]

asked by guest
on Oct 24, 2024 at 10:23 am



Mathbot Says...

I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.