# A Curve Example

 In this example, we will find the area under the curve y = sin(x) from x = 0 to x = .   This is a more complicated than the first examples since we're now trying to fit geometric shapes under a curving line.   This area is the same as We'll be using the "Area" tool so bring that window up.   Graph the curve y = sin(x) in the "Area" tool. This time we'll have to use multiple geometric shapes.   And we'll need to do 2 calculations: One to calculate the positive area which is from x = 0 to x = , and another to calculate the negative area which is from x = 2.5 to x = . Note that is about 4.7, and is the x-coordinate of the lowest point on the graph of y = sin(x) between x = 0 and x = 5. Then we'll account for the negative area by subtracting the area we got for x = 2.5 to x = from the area we got for x = 0 to x = . The steps are: Drop down as many different shapes as it takes to get close to estimating the positive area.   You might start with a trapezoid, but you'll need more shapes.   Try to not overlap the shapes since you'll get a more accurate estimate with minimum overlapping.   Call this area A1. Move all the shapes from part 1 off the screen. Now drop down as many different shapes as it takes to get close to estimating the negative area.   Call this area A2. Then the area under the curve from x = 0 to x = is A1 - A2.   Your answer for A1 - A2 should be very close to 1 since we know that If you experiment a little bit, you'll find that you're better able to approximate the curved line of the graph with the straight lines of the shapes.   So using more shapes yields a better estimate of the area.   Recall that Reimann Sums estimate the area with rectangles of fixed width, and that the integral is then the limit, as the rectangle width goes to 0, of the sum of the area of the rectangles.   Reimann Sums in the limit works because it's like estimating the area with an infinite number of rectangles under the curve.   That is, more shapes means a better estimate, so an infinite number of shapes means the best estimate.