Question

Evaluate the integral ∫2sds limit from a to b?

Answer 1

`b^2-a^2`

Explanation:

`int_a^b2sds`

using the power rule

`intx^ndx=x^(n+1)/(n+1), n!=-1`

`int_a^b2sds=[(2s^2)/2]_a^b`

`=[s^2]_a^b=b^2-a^2`