problems with basic integral

[mcwang719]

hello. this seems like an easy integral i just can't seem to get the right solution which is y^2/4
int y^2/2y dy. can someone help. thanks!!

 

[soroban]

Hello, mcwang719!

\(\displaystyle \int \frac{y^2}{2y}\,dy\)
Answer: \(\displaystyle \frac{y^2}{4}\)

Did you notice that: \(\displaystyle \,\frac{y^2}{2y}\;=\;\frac{y}{2}\;?\)

 

[mcwang719]

thanks soroban!! im so blind.