Homework Help

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

user profile pic

sodelete | Student, College Freshman | (Level 1) Honors

Posted July 18, 2011 at 6:32 PM via web

dislike 0 like

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

2 Answers | Add Yours

Educator Approved

Educator Approved
user profile pic

tonys538 | Student, Undergraduate | TA | (Level 1) Valedictorian

Posted February 28, 2015 at 7:34 PM (Answer #2)

dislike 1 like

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`

user profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted July 18, 2011 at 6:57 PM (Answer #1)

dislike 0 like

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.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes