1st and 2nd Fundamental Theorem of Calculus

Question

1. Consider the function f:[3,6]→Real  defined by f(x)=x².
2. If the domain interval of function f is subdivided into n subintervals of equal width, then the right endpoint of each subinterval is of the form 3+ (3i/n). Explain.
3. Find an explicit expression for an approximation of ∫f(x) dx, from 3 to 6, using a right sum with n subintervals of equal width, and then take a limit as n→∞ to compute an exact evaluation of ∫f(x)dx from 3 to 6.

Answer 1

the interval [3,6] is 6-3 = 3 in length

with n subintervals of equal length each subinterval is 3/n in length

integral of f(x) dx= integral of x^2dx = F(x) = x^3/3 + c where c=an unknown constant

evaluated from 3 to 6 F(x) = 6^3/3 -3^3/3 = 216/3 - 27/3 = 72-9 = 63 square units

using right end heights will overestimate the integral, but in the limit as n approaches infinity, it will get increasingly closer to the actual value.

let n= 3, each rectangle has a base = 3/3 = 1, and heights f(4)= 4^2 =16, f(5) = 5^2= 25 f(6)= 6^2 = 36. sum = 16+25+36 = 80 > 63

let n= 6, each rectangle has base = 3/6 = 1/2, heights: 80 + f(7/2) = 49/4 + f(9/2) = 81/4 + f(11/2) = 121/4 . sum = 80 +12.25 + 20.25 + 30.25 = 80+62.75 = 142.75. then multiply by the base = 1/2 gives 71.375 which is closer to 63