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

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The definite integral `int_0^1 x^2+x dx` has to be determined.

The derivative of x^n is `(x^(n+1))/(n+1)` . Using this rule, the given integral can be written as:

`int_0^1 x^2+x dx`

= `[x^3/3 + x^2/2]_0^1`

= `1/3 - 0 + 1/2 - 0`

= `5/6`

The required definite integral `int_0^1 x^2+x dx = 5/6`

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.**