/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/ 1/1/1/1/1/1ããããããããã / 2· In this blem, we explore two functions that are not presented in a way that you may not be used to ppose f(ax) is defined on the interval (0, 1] by the following rule: we f(x) is the first digit in the decimal expansion for or instance, f(0.719)-7 and f(e-l ) = 3. But note that some numbers have two deci al expansions r instance, 1-0.5-0.499 , , and 2Å 0.35-0.34999 . . .. In such cases, one decinnal expansion s with an infinite trail of 9's and the other terminates. We shall agree to the con will always use the terminating decimal expansion. Hence f(1-5 and f(h) = 3. i. Sketch the graph of y = f(x) with appropriate scales for x and u. ii. Explain why y f(x) is integrable on [0,1]. iii. Calculate the integral f(x) dz. Can you use fundamental theorem of calculus? (b) Suppose g(x) is defined on the interval (0, 1] by the following similar rule. g(x) is the second digit in the decimal expansion forT Again we use the convention that if z has two such expansions, we use the terminating expansion.) For example g(0.719) = 1, g(e-1-6, g() = 0, and gun)-5. i. Explain why y = g(x) is integrable on [0,1]. i. Calculate the integral 10 f() dz. (A graph may help, but is not required.) Can you use the fundamental theorem of calculus? (c) Bonus! It actually turns out that it does not matter which decimal expansion we choose in the definition of either f(x) or g(x). For either convention, we get the same value of the integral. In fact, if a is a number with two decimal expansions for which the first digits differ, we can just leave f(a) undefined. (Similarly, if b is a number with two decimal expansions for which the second digits differ, we can just leave g(a) undefined.) Explain why this is true.