Integral! Help me evaluate this please!

1
∫ (x^2+√x) dx=
0

The antiderivative of x^2 is x^3 / 3 + C
The antiderivative of √x is (2/3) * x^(3/2) + C

∫ (x^2+√x) dx =

x^3 / 3 + (2/3)*x^(3/2)

Evaluate this at 1, then subtract evaluated at 0:

//The exponents don't really matter since it's 0 and 1
[1/3 + (2/3)*1] - [0/3 + (2/3)*0]

//Evaluating at 0 is just 0.

1/3 + 2/3 = 3/3 = 1

Hope that helps :)

1
∫ (x^2+√x) dx= ∫ x^2dx + ∫x^.5dx = (x^3)/3 + (x^1.5)/1.5 +c ] = (1/3+1/1.5) - (0+0) = 1
0

Rule
int of x^n dx = (x^(n+1))/(n+1)