Area under a positive curve can be approximated with rectangles.
Approximate the area under y = 3x+2 between x = 2 and
x = 8 by dividing [2,8] into n = 3 equal parts and
making rectangles, using the left side for the height.
Estimate the area under y = x² + 1 between x = 0 and x = 4 for n = 4.
|f (x) = x² + 1, a = 0, b = 4
n = 4 = number of triangles
We need to add up the area of
4 rectangles (the height of the
fourth --and last-- rectangle
is the y value when x = 3).
Area of 1st rect = (ht) (w) = f (0) x = (0² + 1) (1) = 1
Area of 2nd rect = (ht) (w) = f (1) x = (1² + 1) (1) = 2
Area of 3rd rect = (ht) (w) = f (2) x = (2² + 1) (1) = 5
Area of 4th rect = (ht) (w) = f (3) x = (3² + 1) (1) = 10
Area of all 4 rectangles = 1 + 2 + 5 + 10 = 18 square units
Over a fixed interval [a,b], the more rectangles there are, the
better the approximation will be. The approximation improves as
n gets larger. The actual area under a positive curve between x = a
and x = b would be a limit of the areas of the rectangles as and the width of the rectangles .
This last limit is how we define the definite integral:
The amazing Fundamental Theorem of Calculus says that this limit
ends up being the antiderivative of f(x) evaluated at x = b MINUS the
antiderivative evaluated at x = a. We write:
Area under f(x) > 0 from x = a to x = b is
When determining area, use any antiderivative. Let c = 0 to
make it easy.
Find the exact area under y = x² + 1 from x = 0 to x = 4