Question

How do you find the definite integral of `(sqrt(2x+1))dx` from `[1, 4]`?

Answer 1

It is

`int sqrt(2x+1)dx=1/3*int (2x+1)'*sqrt(2x+1)dx=1/3*[2x+1]^(3/2)+c`

Hence

`int_1^4 sqrt(2x+1)dx=[1/3*[2x+1]^(3/2)]_1^4=9-sqrt(3) ~~ 7.2679`