Calculate the definite integral of y=x^2+x, if x=0 and x=1?

To calculate the definite integral of the given function,
within the given limits of integration, we'll apply Leibniz Newton
formula:

b

`int` f(x)dx = F(b)
-
F(a)

a

1                          
1                 1

`int` (x^2 + x)dx = `int` x^2 dx +
`int` x dx

0                         0                
0

1

`int` (x^2 + x)dx = x^3/3
(x=0 to x=1) + x^2/2 (x=0 to
x=1)

0

1

`int`
(x^2 + x)dx = 1^3/3 - 0^3/3 + 1^2/2 -
0^2/2

0

1

`int`
(x^2 + x)dx = 1/3 +
1/2

0

1

`int`
(x^2 + x)dx =
5/6

0

The value
of the definite integral of the given function, within the limits of integration x = 0
and x = 1, is 5/6.