The first step is to sketch the curve on a graph.
This curve is a parabola, a U-shaped curve in which the arms spread out gradually. The key points of a parabola are intercepts, one at least, and the vertex, which is where the curve changes direction. The vertical line through the vertex is a line of symmetry acting like a mirror reflecting the two halves of the curve. If we rewrite the equation: y-1=3(x-2)^2 we can find the origin of the parabola. When y=1, x=2 which tells us the origin is (2,1), which is the vertex and the line x=2 is the line of symmetry. When x=0, y=13, so the curve cuts the y axis at (0,13). When y=0 (the x axis) the curve has no real value so it doesn't touch or cut the x axis. This is enough to sketch the curve and actually see the area you need to find.
The next step is to enclose the area by establishing limits. The normal limits will trap an area between the curve and one or both axes. These will become the integration limits so that a definite integration can take place.
Now we build the integral by dividing the area into thin rectangular strips. The length of a vertical rectangle is y and its width is dx. The area is ydx and the sum of the areas of the rectangles gives the area under the curve so we have integral(ydx)=integral((3(x-2)^2-1)dx) and the limits for integration will be a to b, a<x<b. integral((3(x-2)^2-1)dx)=
integral((3x^2-12x+12-1)dx)=
[x^3-6x^2+11x](a<x<b)=
(b^3-6b^2+11b)-(a^3-6a^2+11a)=
(b^3-a^3)-6(b^2-6a^2)+11(b-a).
This is the general method. Other areas can be calculated by similarly dividing areas into thin rectangles vertically or horizontally and applying integration. Sometimes it's necessary to subtract one area from another to find another area. Integration will sometimes generate negative areas, but this probably just means that the area is below the axis rather than above it. Note how the lead up to integration is applying simple geometry by dividing up areas into thin rectangles.
