Homework Help

What is the area of the region between x^2-1, x=-2,x=2?

user profile pic

yapayapa | Honors

Posted July 8, 2013 at 4:32 PM via web

dislike 1 like

What is the area of the region between x^2-1, x=-2,x=2?

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 8, 2013 at 4:48 PM (Answer #1)

dislike 1 like

You need to evaluate the area of the region located between the curves `y = x^2 - 1, x = -2, x = 2` , hence, you need to evaluate the following definite integral, such that:

`int_(-2)^2 |f(x)|dx = int_0^2 (f(x) + f(-x))dx`

You need to evaluate `f(x) + f(-x)` such that:

`f(x) + f(-x) = x^2 - 1 + (-x)^2 - 1`

`f(x) + f(-x) = 2x^2 - 2`

`f(x) + f(-x) = 2(x^2 - 1)`

Replacing ` 2(x^2 - 1)` for `f(x) + f(-x)` yields:

`int_(-2)^2 |f(x)|dx = int_0^2 2(x^2 - 1) dx`

Using the property of linearity of integral, yields:

`int_0^2 2(x^2 - 1) dx = 2(int_0^2 x^2 dx - int_0^2 dx)`

`int_0^2 2(x^2 - 1) dx = 2(x^3/3|_0^2 - x|_0^2)`

Using the fundamental theorem of calculus, yields:

`int_0^2 2(x^2 - 1) dx = 2(2^3/3 - 0^3/3 - 2 + 0)`

`int_0^2 2(x^2 - 1) dx = 2(8/3 - 2)`

`int_0^2 2(x^2 - 1) dx = 4/3`

Hence, evaluating the area of the region located between the curves `y = x^2 - 1, x = -2, x = 2` yields `int_0^2 2(x^2 - 1) dx = 4/3.`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes