Integral of 3x^2 -4x -1 = x^3 -2x^2 - x
evaluated from a=2 to b= 6
= 6^3 - 2(6^2) -6 - (2^3 -2(2^2) -2)
= 216 -72- 6 - (8-8-2)
= 138 +2
= 140
or calculate the area under the curve = the integral
approximate,
divide the area into rectangles, base =1 for 4 rectangles
heights using midpoints = f(5/2), f(7/2), f(9/2), f(11/2)
sum of the rectangles' areas =f(5/2) + f(7/2)+f(9/2)+f(11/2) times base = 1
= 3(5/2)^2 -4(5/2)-1 +3(7/2)^2 -4(7/2) -1 +3(9/2)^2 -4(9/2)-1 + 3(11/2)^2 -4(11/2) -1
= 75/4 - 11 + 147/4 - 15 + 243/4 -19 + 484/4 - 23
= 949/4 - 68
= 237.25 -68
= 149.25 which is fairly close to the actual integral of 140 square units
as the number of rectangles increases, approaching infinity, the rectangles' areas approach 140 as a limit
f(x) = y = 3x^2 -4x -1 is an upward opening parabola
f(2) = 12-8-1 = 3
f(6)= 108-24-1 = 83
draw a line from (2,3) to (6,83)
and calculate the area under the rectangle with triangle on top
= 4x2 + .5(4x80) = 8+160= 168
this is an overestimate as the parabola is curved not a straight line
168 is an upper bound for the integral or area