# What is the area enclosed by y+x^2-4x-4 = 0 and the x-axis.

Asked on by lxsptter

### 1 Answer |Add Yours

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The area enclosed by y + x^2 - 4x - 4 = 0 and the x-axis has to be determined.

The point of intersection of the curve with the x-axis is the solution of x^2 - 4x - 4 = 0

=> x1 = `4/2 + sqrt(32)/2`

=> x1 = `2 + sqrt 8`

x2 = `2 - sqrt 8`

The required area is the definite integral

`int_(2 - sqrt 8)^(2 + sqrt 8) -x^2 + 4x + 4 dx`

=> `-x^3/3 + 2x^2 + 4x` between `2 - sqrt 8` and `2 + sqrt 8`

=> `(-1/3)((2 + sqrt 8)^3 - (2 - sqrt 8)^3) + 2((2 + sqrt 8)^2 - (2 - sqrt 8)^2) `

`+ 4((2 + sqrt 8 - 2 + sqrt 8)`

=> `(-1/3)((2 + sqrt 8)^3 - (2 - sqrt 8)^3) + 16*sqrt 8 + 8*sqrt 8`

=> `(-1/3)((2 + sqrt 8)^3 - (2 - sqrt 8)^3) + 24*sqrt 8`

=> `(-1/3)*(8 + 3*4*sqrt 8 + 3*2*8 + (sqrt 8)^3 - 8 - 3*2*8 + 3*4*sqrt 8 + (sqrt 8)^3)`

` + 24*sqrt 8`

=> `(-40*sqrt 8)/3 + 24*sqrt 8`

=> `(32*sqrt 8)/3`

`~~30.169`

The area enclosed by the curve y + x^2 - 4x - 4 = 0 and x-axis is `(32*sqrt 8)/3`

We’ve answered 317,431 questions. We can answer yours, too.