how to solve for the area under the curve

and how to actually graph the equation
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It's best to sketch the curve first. This helps to picture how to find the area under the curve. It's a sideways parabola. In a way, you can swap the x and y variables so that the graph looks like y=x^2-2x but the axes are interchanged to produce x=y^2-2y. When x=0, y^2-2y=0=y(y-2) so y=0 (origin) and 2 are the y intercepts. When y=0, x=0 (origin). The vertex is when 2y-2 (the derivative with respect to y) is zero, so y=1 and x=-1. The vertex is at (-1,1). The second derivative is 2, so the turning point is a minimum. The graph is U-shaped lying on its side with arms pointing to the right (positive). The parabola has its vertex in quadrant 2 and its arms cross the y axis at 0 and 2.

x=y^2-2y, but what is y in terms of x? completing the square we have y^2-2y+1=1+x, (y-1)^2=1+x and y=1+sqrt(1+x). There are generally two values of y for the same value of x. The graph illustrates this, because the parabola is lying on its side so the x value picks up two values of y except at the vertex. We need the lower part of the graph only, given by y=1-sqrt(1+x). If on the graph we paint the inside of the parabola blue and the outside red, this will help us to identify the areas. We want, I think, the red area between the vertex and the x axis and, I assume, up to the y axis, because the curve dips below the x axis when x is positive. Imagine thin rectangles width dx lying vertically with a length equal to y=1-sqrt(1+x). The length of the rectangles for y=1+sqrt(x) include blue and red areas. The area of each rectangle is ydx and the area under the graph is the sum total of the areas of the thin rectangles. When the rectangles are infinitely thin this sum total is the integral of ydx. If we wanted, for example, just the blue area we would need to subtract the red area from the blue-and-red area. This is why it's important to sketch a graph and understand how to apply the mathematics. So our thin rectangles start at x=-1, when y=1, so this rectangle has a width dx and a length of 1; and ends with a rectangle of zero height when x=0. If we integrate between the x limits of -1 and 0 we will get the red area we need: integral((1-sqrt(1+x))dx)=integral((1-(1+x)^1/2)dx)=[x-2(1+x)^(3/2)/3](-1<x<0). We subtract the expression evaluated when x=-1 from the expression when x=0. When x=-1: -1-2(0)^(3/2)/3=-1; when x=0: 0-2(1)^(3/2)/3=-2/3. So the area is -2/3-(-1)=-2/3+1=1/3.

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