# Find the area of the region enclosed by the curves 4x=y^2-4 and 4x=y+16

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### 1 Answer

You need to evaluate definite integral, such that:

`int_a^b |(f(x) - g(x))|dx`

Considering `f(x) = 4x - 16` and `g(x) = sqrt(4x + 4),` you need to find the limits of integration a,b, solving the following equation, such that:

`4x - 16 = sqrt(4x + 4) => 4(x - 4) = sqrt(4x + 4)`

Raising to square both sides to remove the square root, yields:

`16(x - 4)^2 = 4x + 4 => 4(x - 4)^2 = x + 1`

`4x^2 - 32x + 64 = x + 1 => 4x^2 - 33x + 63 = 0`

Using quadratic formula, yields:

`x_(1,2) = (33 +- sqrt(1089 - 1008))/8 => x_(1,2) = (33 +- sqrt(81))/8`

`x_(1,2) = (33 +- 9)/8 => x_1 = 42/8 => x_1 = 21/4`

`x_2 = 24/8 => x_2 = 3`

Hence, evaluating the limits of integration yields `a = 3, b = 21/4.`

`int_3^(21/4)|4x - 16 - sqrt(4x + 4)|dx `

Using the property of linearity of integrals yields:

`int_3^(21/4)(sqrt(4x + 4) - 4x + 16)dx =int_3^(21/4)4x dx - int_3^(21/4) 16dx - int_3^(21/4)(sqrt(4x + 4)) dx`

You may evaluate the integral `int_3^(21/4)(sqrt(4x + 4)) dx` using the following substitution such that:

`x + 1 = t => dx = dt`

`int_3^(21/4)(sqrt(4x + 4)) dx = int_4^(21/4 + 1)2(sqrt t) dt`

`int_4^(21/4 + 1)2(sqrt t) dt = (4/3)t^(3/2)|_4^(21/4 + 1)`

Using the fundamental theorem of calculus yields:

`int_4^(21/4 + 1)2(sqrt t) dt = (4/3)((25/4)^(3/2) - 4^(3/2))`

`int_4^(21/4 + 1)2(sqrt t) dt = (4/3)((25/4)*(5/2) - 4*2)`

`int_4^(21/4 + 1)2(sqrt t) dt = 244/24`

`int_3^(21/4)4x dx = 2x^2|_3^(21/4) => int_3^(21/4)4x dx = 2(441/16 - 9) = 297/8`

`int_3^(21/4)(4x - 16 - sqrt(4x + 4))dx = 297/8 - 9 - 244/24`

`int_3^(21/4)(sqrt(4x + 4) - 4x + 16)dx = 431/24`

**Hence, evaluating the area enclosed by the given curves, yields **`int_3^(21/4)(sqrt(4x + 4) - 4x + 16)dx = 431/24.`