AUC03 Find the area under the curve y = x2 between x = 0 and x = 2.
First, graph the curve. Using the integral approach the area is
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| The area under this curve is less
than the area within a triangle formed with base along the x axis
from 0 to 2, height from y = 0 to 4 and the slant height from the
point (0,0) to (2,4). Such a triangle has area (1/2)2×4
= 4, and as expected is more than the area of
»2.7
computed with the integral. The curve y = x2 goes through the points (0.0), (1,1) and (2,4) so continue this approximation approach by finding the area of this triangle and trapezoid. The area of the triangle is (1/2)1×2 = 1/2. The area of a trapezoid is (1/2)(sum of the opposite faces)(height) which in this case is (1/2)(1 + 4)(1) = 2.5. The sum of these areas is 3, closer to the 2.7 obtained through the integral. |
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If this process were continued with more narrower and narrower trapezoids the area would approach the 2.7 obtained through the integral.
