1. ## Antiderivative check

integral from 0 to 1 (3+xsqrtx)dx

Soo i just need to see if im on the right track
I rewrote it to (3+x.x^-1/2) soo then would the antiderivative be (3x+ (2/5)x^(5/2) )??
obviously then subbing in the x values after but i just want to make sure i got the antiderivative correct.

2. $3+x\sqrt{x}= 3+x\times x^{\frac{1}{2}} = 3+ x^{\frac{3}{2}}$

3. Originally Posted by pickslides
$3+x\sqrt{x}= 3+x\times x^{\frac{1}{2}} = 3+ x^{\frac{3}{2}}$
So 3x+(2/5)x^(5/2) would be correct since taking the derivative of it is 3+x^(3/2). I relized i made a mistake in writing it and put -(1/2).

I finishe the substitutions and got:
[3(1)+(2/5)(1)^(5/2)]-[3(0)+(2/5)(0)^(5/2)]
which is 17/5 if im not mistaken?

4. $\frac{17}{5}$ I agree