Chapter 5 APPLICATIONS OF INTEGRATION Net Area – Revision

In this section we use integrals to find areas of regions that lie between the graphs of two functions. Consider the region that lies between two curves y = f (x) and y = g(x) and between the vertical lines x = a and x = b, where f and g are continuous functions and f (x) ≥ g(x) for all x in [a, b]. a b x y y = g(x) y = f (x) The area A of this region is A = b a f (x) − g(x) dx (5.1) Find the area of the region enclosed by y = x 2 , y = 4 − x 2 , and x = 0.

Problems 1-3 review sums and differences from Section 1.2. This chapter goes forward to integrals and derivatives. 1. If fo, f l , f2, f3, f4 = 0,2,6,12,20, find the differences vj = f,-fj-and the sum of the v's. 0 The differences are vl, v2, VQ, v4 = 2,4,6,8. The sum is 2 +4 +6 + 8 = 20. This equals f4-fo. 2. If vl, v2, VQ, v4 = 3 , 3 , 3 , 3 and fo = 5, find the f's. Show that f4-fo is the sum of the v's. O E a c h n e w f j is fj-1 + v j. So f i = f o + v l = 5 + 3 = 8. Similarly f2 = fl + v 2 = 8 + 3 = 11. Then f3 = 14 and f4 = 17. The sum of the v's is 12. The difference between fiaSt and ffirSt is also 17-5 = 12. 3. (This is Problem 5.1.5) Show that f j = 5 has differences vj = f j-fj-l = d-1.-The formula gives fo = = Then fl = & and f2 = 2-r-l a * Now find the differences: In general fj-fj-1 = ri-'. This is vi. Adding the v's gives the geometric aerie. l + r + r 2 +. .+rn-l. 1-r-1 Its sum is f,-fo = 5-r-1-r-1 4. Suppose v(x) = 2x for 0 < x < 3 and v(x) = 6 for x > 3. Sketch and find the area from 0 t o x under the graph of v(x). m There are really two cases to think about. If x < 3, the shaded triangle with base x and height 2 s has area = :(base) (height) = kx(2x) = x2. If x > 3 the area is that of a triangle plus a rectangle. The triangle has base 3 and height 6 and area 9. The rectangle has base (x-3) and height 6. Total area = 9 + 6(x-3) = 62-9. The area f (x) has a two-part formula: Areas under v(x) give f(x) small triangle L x (2 x) = x 2 2 complete triangle ' 3 (6) = 9 2 add rectangle 6(x-3) = 6x-18 5. Use four rectangles to approximate the area under the curve y = $x2 from x = 0 to x = 4. Then do the same using eight rectangles. The heights of the four rectangles are f (1) = i, f (2) = 2, f (3) = $, f (4) = 8. The width of each rectangle is one. The sum of the four areas is 1. $ + 1 2 + 1 + 1 8 = 15. The sketch shows that the actual area under the curve is less than 15.

The area under a graph often gives useful information.