Draw the graph of y=sinxy=sin⁡x for 0≤x≤π0≤x≤π, plotting points with xx-coordinates at intervals of π6π6, and labeling them O, A, B, C, D, E, FO, A, B, C, D, E, F successively. We start by making a table of values. The values of sin⁡x we seek are given by examining these triangles;[br][br][br][img]https://undergroundmathematics.org/introducing-calculus/r7912/images/img-7912-1.svg[/img][br][br][br]We can now plot these seven points to give us an approximation of the required graph.[br][br][br][img]https://undergroundmathematics.org/introducing-calculus/r7912/images/r7912-curve.svg[/img][br][br][br]Calculate the area of the polygon OABCDEFOABCDEF by dividing it into strips parallel to the y-axis, otherwise.[br][br][br][img]https://undergroundmathematics.org/introducing-calculus/r7912/images/r7912-polygon.svg[/img][br][br][br][br]If we divide the polygon as suggested, we get two triangles and four trapezia. Notice that the shape is symmetrical, so we can just calculate the areas of one triangle and two trapezia and double the result.[br][br][br][img]https://undergroundmathematics.org/introducing-calculus/r7912/images/r7912-3strips.svg[/img][br][br][br]AreahalfArea1Area2Area3⟹Areahalf=12×π6×12=π24=12×π6×(12+3‾√2)=π24(1+3‾√)=12×π6×(3‾√2+1)=π24(3‾√+2)=π24(4+23‾√)Area1=12×π6×12=π24Area2=12×π6×(12+32)=π24(1+3)Area3=12×π6×(32+1)=π24(3+2)⟹Areahalf=π24(4+23)So the area of the whole polygon is π6(2+3‾√)=1.954π6(2+3)=1.954 (44 s.f.).[br][br]Hence give an approximation to∫π/20sinxdx.∫0π/2sin⁡xdx. The value of the integral we are to approximate is the area under the curve as far as its maximum point. This is approximately the same as half the area of the polygon, 0.9770.977 (33 s.f.). From the diagram, we can see that this is a slight underestimate. [br][br][br]Using calculus, it is possible to calculate the required integral exactly, as −cos(π/2)−(−cos0)=1−cos⁡(π/2)−(−cos⁡0)=1.