Two chemists created a new way to convert unobtainium [Uo] into canotainium [Cn]. After many experiments, they gathered some data about the probability of producing specific amounts of [Cn] at the end of one run of the conversion process. This is their data: r = dp d[Cn] and [Cn] (in mg) 0 2 7 10 r 0.07 0.15 0.12 0.29 Here: p([Cn]) is the probability of ending up with [Cn] mg of canobtainium after one run of the conversion process. The chemists also know that: There is 0.06 probability of finishing the conversion process with no [Cn], and There is a 0.65 probability of finishing the conversion process with 10mg of [Cn]. They need to compute the expected production of [Cn] over one run of the conversion process. Because the rate of change is continuous (even though the data they gathered is not), they decide to use a formula they found in an old dusty statistics textbook: expected value = b 1 a b [*[Cn] p([Cn]) d[Cn], which in this case becomes 10 1 expected value = 10 [*[Cn] p[([Cn]) d[Cn]. Help the scientists estimate this integral. (a) First, the units of the expected value are mg To do this: You might want to change the input variable from [Cn] to x to make it easier to parse the integrals; (a) First, the units of the expected value are mg To do this: You might want to change the input variable from [Cn] to x to make it easier to parse the integrals; Use integration by parts to simplify the formula for the expected value so that you can use the information on the table above; Use a left-hand Riemann sum on the remaining integral with 3 intervals to compute it. (b) Expected value of [Cn] mg