Question

Let g(x) = 1- x and f(x) = x2 - 2x + 1 for Problems 5 and 6 below. 5. (a) Draw a graph of f(x) and g(x) on the same axes, and label their points of intersection. Calculate the area below g(2) and above the z-axis between I = 0 and = 1 using geometric methods. How does the area under f between x = 0 and x = 1 compare to the area under g? (b) Use 4 rectangles to calculate L4, the area estimate under f(2) between = 0 and I = 1 using left endpoints. Then calculate R4, the area estimate using right endpoints. (c) If A is the true area under f(x) between x = 0 and I = 1, compare L4, R4, and A. 6. We will use the following definition to calculate the exact area under f(x) from x = 0 to x = 1. Definition: The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: Area = A = lim Rn = lim [f(x1) Ax + f(22)Ar + ... + f(n)Ar] 100 000 (a) Use the definition of area given above to write a limit that is equal to the area under the graph of f from x = 0 to x = 1. Do not evaluate the limit yet. (b) Evaluate the limit in part (a) using these two formulas, both found in Theorem 3 of Appendix F of your book. 12 + 22 + 32 + ... + n2 = 2 n(n+1) (2n +1) n(n+1) 1+2+3+...+ (c) The answer for part (b) should represent the area under f(x) between I = 0 and 3 = 1. Explain briefly why this limit represents the area under f.

Just need the answer to question 6 using the information provided in the block above question 5. Please be clear due to this being a multi-step problem. Thanks   

Answer 1

f(x) = x² 2x+1 @ Area between x=0 and X=1 Step size 4x= 1-0 = 1 1 1 x=0+ 1 / 1X2 = 0+3 , . .,86 = h . . , x n = 1. Alea A= Lim f(x) axtf(X2) AX + ... + f(xn) Ax] nto kiert Foxx) no hein + 3 () ?24+1) - here (22 an** - hier > ŽK+nº)

Use n(n+1) (2n+1) I k² k=1 K- Alla Athing to (n)/2 +20 () (na) - n°) - lin n(n+1) (2n+1) Lim n(n+1) + 1 nto 6 n3 nto n3 = 2 -1+1 = t3 Alea (A = 1 © This represents alea because as n o , the step size becomes smaller hence the width of the rectangles the Considered for apploximating area becomes smaller Due to this elol in applanimation diminishes and becomes zero as n o . Vecification - 5'(-dx = (x-1) Vecidir til f(x) dx = 16²_2X+1) dx cs