`y = x^2 - 2x, y = x + 4` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`. Then find the...

Since the curves are `y = x^2 - 2x` and y = x + 4, you must integrate y with respect to x. First, you need to find the point of intersection between the curves `y = x^2 - 2x` and y = x + 4, by solving the equation:

`x^2 - 2x = x + 4 => x^2 - 3x - 4 = 0 => x_(1,2) = (3+-sqrt(9+16))/2`

`x_(1,2) = (3+-5)/2 => x_1 = 4,...

Since the curves are `y = x^2 - 2x` and y = x + 4, you must integrate y with respect to x. First, you need to find the point of intersection between the curves `y = x^2 - 2x` and y = x + 4, by solving the equation:

`x^2 - 2x = x + 4 => x^2 - 3x - 4 = 0 => x_(1,2) = (3+-sqrt(9+16))/2`

`x_(1,2) = (3+-5)/2 => x_1 = 4, x_2 = -1`

Hence, x = -1 and x = 4 and these values are the endpoints of the definite integral you need to evaluate to find the area enclosed by the given curves.

You must check what curve is greater than the other on interval `[-1,4]` , hence, you need to check the monotony of the function `f(x) = x^2 - 3x - 4` and you need to notice that f(x) increases on `[3/2,4] ` and it decreases on `[-1,3/2]` , hence, the function `x^2 - 2x > x + 4` on `[3/2,4` ]  and `x^2 - 2x < x + 4` on `[-1,3/2].`

You may evaluate the area such that:

`int_(-1)^4 |x^2 - 3x - 4|dx = int_(-1)^(3/2) (-x^2 + 3x + 4)dx + int_(3/2)^4 (x^2 - 3x - 4) dx`

`int_(-1)^4 |x^2 - 3x - 4|dx = int_(-1)^(3/2) (-x^2)dx + 3int_(-1)^(3/2) xdx + int_(-1)^(3/2) 4dx - int_(3/2)^4 (x^2)dx - 3int_(3/2)^4 xdx - int_(3/2)^4 4dx`

`int_(-1)^4 |x^2 - 3x - 4|dx = (-x^3)/3|_(-1)^(3/2) + 3(x^2)/2|_(-1)^(3/2) + 4x|_(-1)^(3/2) + (x^3)/3|_(3/2)^4 - 3(x^2)/2|_(3/2)^4 - 4x|_(3/2)^4`

`int_(-1)^4 |x^2 - 3x - 4|dx =- 9/8 -1/3 + 27/8 - 3/2 + 6 + 4 + 64/3 - 9/8 +24 + 27/8 - 16 + 6`

`int_(-1)^4 |x^2 - 3x - 4|dx = -18/8 + 54/8 + 63/3 + 24 - 3/2`

`int_(-1)^4 |x^2 - 3x - 4|dx = 36/8 + 21 + 24 - 3/2`

`int_(-1)^4 |x^2 - 3x - 4|dx = 6/2 + 21 + 24`

`int_(-1)^4 |x^2 - 3x - 4|dx =48`

Hence, evaluating the area enclosed by the curves yields `int_(-1)^4 |x^2 - 3x - 4|dx = 48`

The area evaluated above is the area of the region between the red curve and orange curve, for `x in [-1,4].`