SECTION 14.5 The Area Problem; The Integral 969 For a constant function ( ) = f x k and for a linear function ( ) = + f x mx B, we can solve the area problem using formulas from geometry. See Figures 22(a) and (b). Figure 22 f(x) 5 k b 2 a f(x) 5 mx 1 B f(b) f(a) x x k k y y a b A A a b (b) (a) b 2 a A 5 area of rectangle 5 length 3 width 5 k(b 2 a) A 5 area of trapezoid 5 [f(b) 1 f(a)](b 2 a) 2 1 Figure 23 y 5 f (x) y 5 f (x) f(u4) x x y y b 2 a 2 a 5 u1 a u 2 (b) 4 subintervals (a) 2 subintervals b 2 a 2 b b 2 a 4 b 2 a 4 b 2 a 4 b 2 a 4 u1 u3 b 5 u4 u2 f(u1) f(u2) f(u1) f(u2) f(u3) For most other functions, no formulas from geometry are available. We begin by discussing a way to approximate the area under the graph of a function f from a to b. 1 Approximate the Area under the Graph of a Function We use rectangles to approximate the area under the graph of a function f. We do this by partitioning or dividing the interval [ ] a b , into subintervals of equal width. On each subinterval, we form a rectangle whose base is the width of the subinterval and whose length is ( ) f u for some number u in the subinterval. Look at Figure 23. In Figure 23(a), the interval [ ] a b , is partitioned into two subintervals, each of width −b a 2 , and the number u is chosen as the left endpoint of each subinterval. In Figure 23(b), the interval [ ] a b , is partitioned into four subintervals, each of width −b a 4 , and the number u is chosen as the right endpoint of each subinterval. The area A under f from a to b is approximated by adding the areas of the rectangles formed by the partition. Using Figure 23(a), ( ) ( ) ≈ + = − + − A f u b a f u b a Area area of first rectangle area of second rectangle 2 2 1 2 Using Figure 23(b), ( ) ( ) ( ) ( ) ≈ + + + = − + − + − + − A f u b a f u b a f u b a f u b a Area area of first rectangle area of second rectangle area of third rectangle area of fourth rectangle 4 4 4 4 1 2 3 4
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