2. (10 pts) Consider the function f (x) = v9-x. (a) (3 pts) Sketch the the curves f (x), x =5, and y = 0. Shade the region enclosed by these curves. 3 (b) (3 pts) Using n = 4 subintervals, sketch the rectangles for a Riemann sum that will overestimate [v9-xdx.(c) (4 pts) Express the integral 9-xdx as a limit of Riemann sums; indicate whether you are using right-hand or left-hand endpoints.terms. Let's introduce some new notation, called the Riemann Sum, to condense writing out these what we sum E means add terms together Ei f( xi ) Ax summation notation i = I ton are known as our = [fix, ) AX + f ( x 2 ) Axt indices. The first value we plug f ( x ; ) 4x + f ( x4 ) axt ... in for i is i =1 and the last is ien f ( xn ) X ] Z i = , f ( xi ) Ax = f ( x ) AX + f ( xz)ox+ .. . + f (xn)4x Let's rewrite R. and L, in terms of this new notation. Ln = AX f ( x. ) + ex f (x, ) + .. .+ axf(xx-1) : V f ( X;-, )4X Rn = AX f ( x , ) + 4 x f ( x 2 )+ ... tax f(an) f ( xi ) AXExample 3 Area under f(x)= 2+cos(x) on I = [D,n], n= 4 Estimate the area under the graph of /(x) - 2 + cos(x) from = = 0 to = = # using 4 sub-intervals. (a) right endpoints. y= f( x ) width of rectangles = ax = 6-a _ I- O (1/4, f (W/4])= ("/4, 2+72/5) n 4 4 (/2, f ("/2))- (" / z, 2+0) 1( 30 / 4, f(30/6 ) ) - (" /4, 2+ "() Al AL ( IT , f Cal) - (1, 2-1) A3 A4 R4 = At Az + A3 + A4 2 4 = # ( 2+ 2 ) + 4 ( 2) + 4 ( 2- 2 ) +7. 1 =4 [ 2+ 2 + 2 + 2 - 1 2 + 1 ] = 4 (7) 2 5. 49 8 (b) left endpoints. ( 8, 2+ # ) ( 7 , 2 ) LA = Alt Act As + A4 A At ( 31 , 2 -12 ) =# (3 ) + # ( 2+ 2 ) + # ( 2) +# (2- 12) As At = 4 [ 3+2+ 12 + 2+ 2 - 7 ] 0 4 311 IT = " (9) x 7.069 La > area under 2 R4 f 7.069 2 area under f 25.4982. When will our approximation overestimate the area under the curve? Assume fzo on our inenal If f is increasing on an interval If t is decreasing on an Rn will overestimate area under the interval in will overestimate curve the area under the curve