We will apply the foil method here first before integrating it.

Kindly recall that we can rewrite `(1 - y^2)^2` as `(1 - y^2)(1 - y^2)` .

FOIL means first terms, outer terms, inner terms, last terms.

So, we will have:

First terms `= 1 * 1 = 1` .

Outer terms `= 1 * - y^2 = -y^2` .

Inner terms `= -y^2 * 1 = -y^2` .

Last terms` = -y^2 * -y^2 = y^4` .

`(1 - y^2)(1 - y^2) = 1 - y^2 - y^2 + y^4`

Combine like terms.

`(1 - y^2)(1 - y^2) = 1 - y^2 - y^2 + y^4 = 1 - 2y^2 + y^4`

We can now rewrite the problem as` int_-1^1(1 - 2y^2 + y^4)dy` .

We can apply the formula:

`int_a^bu^ndu = (u^{n+1})/(n+1)|_a^b`

So, we will have:

`int_-1^1(1 - 2y^2 + y^4)dy = (y - (2y^3)/(3) + (y^5)/(5))|_-1^1`

Apply the formula: `int_a^bf(x)dx = F(b) - F(a)` .

`(y - (2y^3)/(3) + (y^5)/(5))|_-1^1 = [((1) - (2(1)^3)/(3) + ((1)^5)/(5))] - [((-1) - (2(-1)^3)/(3) + ((-1)^5)/(5))] `

`= [(1 - (2(1))/(3) + (1)/(5)] - [((-1) - (2(-1))/(3) + (-1)/(5))] `

`= [(1) - (2)/(3) + (1)/(5)] - [((-1) - (-2)/(3) + (-1)/(5)] `

Take note that negative times negative is positive.

`[(1) - (2)/(3) + (1)/(5)] - [((-1) - (-2)/(3) + (-1)/(5)]= [(1) - (2)/(3) + (1)/(5)] - [(-1) + (2)/(3) - (1)/(5)] `

Find the lcd, note that identifying the lcd is like finding the least common multiple of

the denominator.

The denominators are `1` , `3` , and `5` .

Multiples of the denominators are:

`1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, ..`

`3,6,9,12,15,18,21, ..`

`5,10,15,20,25, ...`

Therefore, the lcd is `15` .

We will make the denominators equal to `15` .

`[(1) - (2)/(3) + (1)/(5)] - [(-1) + (2)/(3) - (1)/(5)] = [(15)/(15) - (2 * 5)/(15) + (1*3)/(15)] - [(-15)/(15) + (2*5)/(15) - (1*3)/(15)] `

Since` 3 * 5 = 15` , we multiply `(2)/(3)` by `5` on top and bottom. Since `5 *3 = 15` , we multiply `(1)/(5)` by `3` on top and bottom.

`[(15)/(15) - (2 * 5)/(15) + (1*3)/(15)] - [(-15)/(15) + (2*5)/(15) - (1*3)/(15)] = (15 - 10 + 3)/(15) - (-15 + 10 - 3)/(15) = (8)/(15) - (-8)/(15)`

Since, negative times negative is positive we will have:

`(8)/(15) - (-8)/(15) = (8)/(15) + (8)/(15) `

`= (16)/(15).`

Multiply the `pi` outside the integral, the answer will be `(16pi)/(15)` .

That is it! :)

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now